problem solving quadratic functions

Quadratic Functions Problems with Solutions

quadratic functions problems with detailed solutions are presented along with graphical interpretations of the solutions.

Review Vertex and Discriminant of Quadratic Functions

f(x) = a x 2 + b x + c

If a > 0 , the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0 , the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2 + b x + c can be written in vertex form as follows:

f(x) = a (x - h) 2 + k

Problems with Solutions

Problem 1 The profit (in thousands of dollars) of a company is given by.

P(x) = 5000 + 1000 x - 5 x 2

• Function P that gives the profit is a quadratic function with the leading coefficient a = - 5. This function (profit) has a maximum value at x = h = - b / (2a) x = h = -1000 / (2(-5)) = 100
• The maximum profit Pmax, when x = 100 thousands is spent on advertising, is given by the maximum value of function P k = c - b 2 / (4 a) b)
• The maximum profit Pmax, when x = 100 thousands is spent on advertising, is also given by P(h = 100) P(100) = 5000 + 1000 (100) - 5 (100) 2 = 55000.
• When the company spends 100 thousands dollars on advertising, the profit is maximum and equals 55000 dollars.

Problem 2 An object is thrown vertically upward with an initial velocity of V o feet/sec. Its distance S(t), in feet, above ground is given by

• S(t) is a quadratic function and the maximum value of S(t)is given by k = c - b 2 /(4 a) = 0 - (v o ) 2 / (4(-16))
• This maximum value of S(t) has to be 300 feet in order for the object to reach a maximum distance above ground of 300 feet. - (v o ) 2 / (4(-16)) = 300
• we now solve - (v o ) 2 / (4(-16)) = 300 for v o v o = 64*300 = 80 √3 feet/sec.

Problem 3 Find the equation of the quadratic function f whose graph passes through the point (2 , -8) and has x intercepts at (1 , 0) and (-2 , 0). Solution to Problem 3

• Since the graph has x intercepts at (1 , 0) and (-2 , 0), the function has zeros at x = 1 and x = - 2 and may be written as follows. f(x) = a (x - 1)(x + 2)
• The graph of f passes through the point (2 , -8), it follows that f(2) = - 8
• which leads to - 8 = a (2 - 1)(2 + 2)
• expand the right side of the above equation and group like terms -8 = 4 a
• Solve the above equation for a to obtain a = - 2
• The equation of f is given by f(x) = - 2 (x - 1)(x + 2)
• Check answer f(1) = 0 f(-2) = 0 f(2) = - 2 (2 - 1)(2 + 2) = -8

Problem 4 Find values of the parameter m so that the graph of the quadratic function f given by

• To find the points of intersection, you need to solve the system of equations y = x 2 + x + 1 y = m x
• Substitute m x for y in the first equation to obtain mx = x 2 + x + 1
• Write the above quadratic equation in standard form. x 2 + x (1 - m) + 1 = 0
• Find the discriminant D of the above equation. D = (1 - m) 2 - 4(1)(1) D = (1 - m) 2 - 4 a)
• For the graph of f and that of the line to have 2 points of intersection, D must be positive, which leads to (1 - m) 2 - 4 > 0
• Solve the above inequality to obtain solution set for m in the intervals (- ∞ , -1) U (3 , + ∞) b)
• For the graph of f and that of the line to have 1 point of intersection, D must be zero, which leads to (1 - m) 2 - 4 = 0
• Solve the above equation to obtain 2 solutions for m. m = -1 m = 3 c)
• For the graph of f and that of the line to have no points of intersection, D must be negative, which leads to (1 - m) 2 - 4 < 0

Problem 5 The quadratic function C(x) = a x 2 + b x + c represents the cost, in thousands of Dollars, of producing x items. C(x) has a minimum value of 120 thousands for x = 2000 and the fixed cost is equal to 200 thousands. Find the coefficients a,b and c. Solution to Problem 5

• Function C is a quadratic function. Its minimum point, which is given as (2000,120) is the vertex of the graph of C. Hence we can write C(x) in vertex form as follows C(x) = a (x - 2000) 2 + 120
• The fixed cost is the value of C(x) when x = 0. Hence C(0) = a (0 - 2000) 2 + 120 = 200
• Solve for a a = 80 / 2000 2 = 0.00002
• We expand C(x) and identify the coefficients a, b and c. C(x) = 0.00002 (x - 2000) 2 + 120 = 0.00002 x 2 - 0.08 x + 200 a = 0.00002 , b = -0.08 and c = 200.

Problem 6 Find the equation of the tangent line to the the graph of f(x) = - x 2 + x - 2 at x = 1. Solution to Problem 6

• There are at least two methods to solve the above question. Method 1
• Let the equation of the tangent line be of the form y = m x + b
• and we therefore need to find m and b. The tangent line passes through the point (1 , f(1)) = (1 , -2)
• Hence the equation in m and b - 2 = m (1) + b or m + b = -2
• To find the point of tangency of the line and the graph of the quadratic function, we need to solve the system y = m x + b and y = - x 2 + x - 2
• Substitute y by m x + b in the second equation of the system to obtain m x + b = - x 2 + x - 2
• Write the above equation in standard form - x 2 + x (1 - m) - 2 - b = 0
• For the line to be tangent to the graph of the quadratic function, the discriminant D of the above equation must be equal to zero. Hence D = b 2 - 4 a c = (1 - m) 2 - 4 (-1) (- 2 - b) = 0
• which gives (1 - (- 2 - b) ) 2 + 4 (- 2 - b) = 0
• Expand, simplify and write the above equation in standard form b 2 2 b + 1 = 0 (b + 1) 2 = 0
• Solve for b b = - 1
• Find m m = - 2 - b = -1

Questions with Solutions

Find the equation of the quadratic function f whose graph has x intercepts at (-1 , 0) and (3 , 0) and a y intercept at (0 , -4).

Question 2 Find values of the parameter c so that the graphs of the quadratic function f given by f(x) = x 2 + x + c and the graph of the line whose equation is given by y = 2 x have: a) 2 points of intersection, b) 1 point of intersection, c) no points of intersection.

Solutions to the Above Questions

Solution to Question 1

• The x intercepts of the graph of f are the zero of f(x). Hence f(x) is of the form f(x) = a (x + 1)(x - 3)
• We now need to find coefficient a using the y intercept f(0) = a(0 + 1)(0 - 3) = - 4
• Solve for a a = 4 / 3
• Hence f(x) = (4 / 3) (x + 1)(x - 3)

Solution to Question 2

• To find the coordinates of the point of intersections of the graphs of f(x) = x 2 + x + c and y = 2 x, we need to solve the system y = x 2 + x + c and y = 2 x
• which by substitution , gives the equation x 2 + x + c = 2x
• Rewrite the above equation in standard form x 2 - x + c = 0
• Find the discriminant D D = 1 - 4 c
• Conclusion If D is positive or c < 1 / 4 , the two graphs intersect at two points. If D is equal to 0 or c = 1 / 4 , the two graphs intersect (touch) at 1 point. If D is negative or c > 1 / 4 , the two graphs have no point of intersection.

More References and Links to Quadratic Functions

• Math Questions With Answers (13): Quadratic Functions.
• Vertex and Intercepts Parabola Problems.
• Find Vertex and Intercepts of Quadratic Functions - Calculator: An applet to solve calculate the vertex and x and y intercepts of the graph of a quadratic function.
• Quadratic functions (general form).
• Quadratic functions (standard form).
• Graphing quadratic functions.
• Solver to Analyze and Graph a Quadratic Function

5.1 Quadratic Functions

Learning objectives.

In this section, you will:

• Recognize characteristics of parabolas.
• Understand how the graph of a parabola is related to its quadratic function.
• Determine a quadratic function’s minimum or maximum value.
• Solve problems involving a quadratic function’s minimum or maximum value.

Curved antennas, such as the ones shown in Figure 1 , are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.

In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.

Recognizing Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola . One important feature of the graph is that it has an extreme point, called the vertex . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry . These features are illustrated in Figure 2 .

The y -intercept is the point at which the parabola crosses the y -axis. The x -intercepts are the points at which the parabola crosses the x -axis. If they exist, the x -intercepts represent the zeros , or roots , of the quadratic function, the values of x x at which y = 0. y = 0.

Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y - y - intercept of the parabola shown in Figure 3 .

The vertex is the turning point of the graph. We can see that the vertex is at ( 3 , 1 ) . ( 3 , 1 ) . Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. x = 3. This parabola does not cross the x - x - axis, so it has no zeros. It crosses the y - y - axis at ( 0 , 7 ) ( 0 , 7 ) so this is the y -intercept.

Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions

The general form of a quadratic function presents the function in the form

where a , b , a , b , and c c are real numbers and a ≠ 0. a ≠ 0. If a > 0 , a > 0 , the parabola opens upward. If a < 0 , a < 0 , the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by x = − b 2 a . x = − b 2 a . If we use the quadratic formula, x = − b ± b 2 − 4 a c 2 a , x = − b ± b 2 − 4 a c 2 a , to solve a x 2 + b x + c = 0 a x 2 + b x + c = 0 for the x - x - intercepts, or zeros, we find the value of x x halfway between them is always x = − b 2 a , x = − b 2 a , the equation for the axis of symmetry.

Figure 4 represents the graph of the quadratic function written in general form as y = x 2 + 4 x + 3. y = x 2 + 4 x + 3. In this form, a = 1 , b = 4 , a = 1 , b = 4 , and c = 3. c = 3. Because a > 0 , a > 0 , the parabola opens upward. The axis of symmetry is x = − 4 2 ( 1 ) = −2. x = − 4 2 ( 1 ) = −2. This also makes sense because we can see from the graph that the vertical line x = −2 x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, ( −2 , −1 ) . ( −2 , −1 ) . The x - x - intercepts, those points where the parabola crosses the x - x - axis, occur at ( −3 , 0 ) ( −3 , 0 ) and ( −1 , 0 ) . ( −1 , 0 ) .

The standard form of a quadratic function presents the function in the form

where ( h , k ) ( h , k ) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function .

As with the general form, if a > 0 , a > 0 , the parabola opens upward and the vertex is a minimum. If a < 0 , a < 0 , the parabola opens downward, and the vertex is a maximum. Figure 5 represents the graph of the quadratic function written in standard form as y = −3 ( x + 2 ) 2 + 4. y = −3 ( x + 2 ) 2 + 4. Since x – h = x + 2 x – h = x + 2 in this example, h = –2. h = –2. In this form, a = −3 , h = −2 , a = −3 , h = −2 , and k = 4. k = 4. Because a < 0 , a < 0 , the parabola opens downward. The vertex is at ( − 2 , 4 ) . ( − 2 , 4 ) .

The standard form is useful for determining how the graph is transformed from the graph of y = x 2 . y = x 2 . Figure 6 is the graph of this basic function.

If k > 0 , k > 0 , the graph shifts upward, whereas if k < 0 , k < 0 , the graph shifts downward. In Figure 5 , k > 0 , k > 0 , so the graph is shifted 4 units upward. If h > 0 , h > 0 , the graph shifts toward the right and if h < 0 , h < 0 , the graph shifts to the left. In Figure 5 , h < 0 , h < 0 , so the graph is shifted 2 units to the left. The magnitude of a a indicates the stretch of the graph. If | a | > 1 , | a | > 1 , the point associated with a particular x - x - value shifts farther from the x- axis, so the graph appears to become narrower, and there is a vertical stretch. But if | a | < 1 , | a | < 1 , the point associated with a particular x - x - value shifts closer to the x- axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5 , | a | > 1 , | a | > 1 , so the graph becomes narrower.

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

For the linear terms to be equal, the coefficients must be equal.

This is the axis of symmetry we defined earlier. Setting the constant terms equal:

In practice, though, it is usually easier to remember that k is the output value of the function when the input is h , h , so f ( h ) = k . f ( h ) = k .

Forms of Quadratic Functions

A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola.

The general form of a quadratic function is f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c where a , b , a , b , and c c are real numbers and a ≠ 0. a ≠ 0.

The standard form of a quadratic function is f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k where a ≠ 0. a ≠ 0.

The vertex ( h , k ) ( h , k ) is located at

Given a graph of a quadratic function, write the equation of the function in general form.

• Identify the horizontal shift of the parabola; this value is h . h . Identify the vertical shift of the parabola; this value is k . k .
• Substitute the values of the horizontal and vertical shift for h h and k . k . in the function f ( x ) = a ( x – h ) 2 + k . f ( x ) = a ( x – h ) 2 + k .
• Substitute the values of any point, other than the vertex, on the graph of the parabola for x x and f ( x ) . f ( x ) .
• Solve for the stretch factor, | a | . | a | .
• Expand and simplify to write in general form.

Writing the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function g g in Figure 7 as a transformation of f ( x ) = x 2 , f ( x ) = x 2 , and then expand the formula, and simplify terms to write the equation in general form.

We can see the graph of g is the graph of f ( x ) = x 2 f ( x ) = x 2 shifted to the left 2 and down 3, giving a formula in the form g ( x ) = a ( x − ( −2 ) ) 2 − 3 = a ( x + 2 ) 2 – 3. g ( x ) = a ( x − ( −2 ) ) 2 − 3 = a ( x + 2 ) 2 – 3.

Substituting the coordinates of a point on the curve, such as ( 0 , −1 ) , ( 0 , −1 ) , we can solve for the stretch factor.

In standard form, the algebraic model for this graph is ( g ) x = 1 2 ( x + 2 ) 2 – 3. ( g ) x = 1 2 ( x + 2 ) 2 – 3.

To write this in general polynomial form, we can expand the formula and simplify terms.

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

We can check our work using the table feature on a graphing utility. First enter Y1 = 1 2 ( x + 2 ) 2 − 3. Y1 = 1 2 ( x + 2 ) 2 − 3. Next, select TBLSET, TBLSET, then use TblStart = – 6 TblStart = – 6 and Δ Tbl = 2, Δ Tbl = 2, and select TABLE . TABLE . See Table 1 .

The ordered pairs in the table correspond to points on the graph.

A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 8 . Assume that the point (–4, 7) is the highest point of the basketball’s trajectory. Find an equation for the path of the ball. Does the shooter make the basket?

Given a quadratic function in general form, find the vertex of the parabola.

• Identify a ,   b ,   and   c . a ,   b ,   and   c .
• Find h , h , the x -coordinate of the vertex, by substituting a a and b b into h = – b 2 a . h = – b 2 a .
• Find k , k , the y -coordinate of the vertex, by evaluating k = f ( h ) = f ( − b 2 a ) . k = f ( h ) = f ( − b 2 a ) .

Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function f ( x ) = 2 x 2 – 6 x + 7. f ( x ) = 2 x 2 – 6 x + 7. Rewrite the quadratic in standard form (vertex form).

The horizontal coordinate of the vertex will be at h = − b 2 a = − −6 2 ( 2 ) = 6 4 = 3 2 The vertical coordinate of the vertex will be at k = f ( h ) = f ( 3 2 ) = 2 ( 3 2 ) 2 − 6 ( 3 2 ) + 7 = 5 2 The horizontal coordinate of the vertex will be at h = − b 2 a = − −6 2 ( 2 ) = 6 4 = 3 2 The vertical coordinate of the vertex will be at k = f ( h ) = f ( 3 2 ) = 2 ( 3 2 ) 2 − 6 ( 3 2 ) + 7 = 5 2

Rewriting into standard form, the stretch factor will be the same as the a a in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the " a a " from the general form.

The standard form of a quadratic function prior to writing the function then becomes the following:

One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, k , k , and where it occurs, x . x .

Given the equation g ( x ) = 13 + x 2 − 6 x , g ( x ) = 13 + x 2 − 6 x , write the equation in general form and then in standard form.

Finding the Domain and Range of a Quadratic Function

Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y -values greater than or equal to the y -coordinate at the turning point or less than or equal to the y -coordinate at the turning point, depending on whether the parabola opens up or down.

Domain and Range of a Quadratic Function

The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions.

The range of a quadratic function written in general form f ( x ) = a x 2 + b x + c f ( x ) = a x 2 + b x + c with a positive a a value is f ( x ) ≥ f ( − b 2 a ) , f ( x ) ≥ f ( − b 2 a ) , or [ f ( − b 2 a ) , ∞ ) ; [ f ( − b 2 a ) , ∞ ) ; the range of a quadratic function written in general form with a negative a a value is f ( x ) ≤ f ( − b 2 a ) , f ( x ) ≤ f ( − b 2 a ) , or ( − ∞ , f ( − b 2 a ) ] . ( − ∞ , f ( − b 2 a ) ] .

The range of a quadratic function written in standard form f ( x ) = a ( x − h ) 2 + k f ( x ) = a ( x − h ) 2 + k with a positive a a value is f ( x ) ≥ k ; f ( x ) ≥ k ; the range of a quadratic function written in standard form with a negative a a value is f ( x ) ≤ k . f ( x ) ≤ k .

Given a quadratic function, find the domain and range.

• Identify the domain of any quadratic function as all real numbers.
• Determine whether a a is positive or negative. If a a is positive, the parabola has a minimum. If a a is negative, the parabola has a maximum.
• Determine the maximum or minimum value of the parabola, k . k .
• If the parabola has a minimum, the range is given by f ( x ) ≥ k , f ( x ) ≥ k , or [ k , ∞ ) . [ k , ∞ ) . If the parabola has a maximum, the range is given by f ( x ) ≤ k , f ( x ) ≤ k , or ( − ∞ , k ] . ( − ∞ , k ] .

Find the domain and range of f ( x ) = − 5 x 2 + 9 x − 1. f ( x ) = − 5 x 2 + 9 x − 1.

As with any quadratic function, the domain is all real numbers.

Because a a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x - x - value of the vertex.

The maximum value is given by f ( h ) . f ( h ) .

The range is f ( x ) ≤ 61 20 , f ( x ) ≤ 61 20 , or ( − ∞ , 61 20 ] . ( − ∞ , 61 20 ] .

Find the domain and range of f ( x ) = 2 ( x − 4 7 ) 2 + 8 11 . f ( x ) = 2 ( x − 4 7 ) 2 + 8 11 .

Determining the Maximum and Minimum Values of Quadratic Functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola . We can see the maximum and minimum values in Figure 9 .

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

• ⓐ Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L . L .
• ⓑ What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W , W , to represent the width of the garden and the length of the fence section parallel to the backyard fence.

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

This formula represents the area of the fence in terms of the variable length L . L . The function, written in general form, is

• ⓑ The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a a is the coefficient of the squared term, a = −2 , b = 80 , a = −2 , b = 80 , and c = 0. c = 0.

To find the vertex:

The maximum value of the function is an area of 800 square feet, which occurs when L = 20 L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11 .

Given an application involving revenue, use a quadratic equation to find the maximum.

• Write a quadratic equation for a revenue function.
• Find the vertex of the quadratic equation.
• Determine the y -value of the vertex.

Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p p for price per subscription and Q Q for quantity, giving us the equation Revenue = p Q . Revenue = p Q .

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 p = 30 and Q = 84,000. Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 p = 32 and Q = 79,000. Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y -intercept.

This gives us the linear equation Q = −2,500 p + 159,000 Q = −2,500 p + 159,000 relating cost and subscribers. We now return to our revenue equation.

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

This could also be solved by graphing the quadratic as in Figure 12 . We can see the maximum revenue on a graph of the quadratic function.

Finding the x - and y -Intercepts of a Quadratic Function

Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y - y - intercept of a quadratic by evaluating the function at an input of zero, and we find the x - x - intercepts at locations where the output is zero. Notice in Figure 13 that the number of x - x - intercepts can vary depending upon the location of the graph.

Given a quadratic function f ( x ) , f ( x ) , find the y - y - and x -intercepts.

• Evaluate f ( 0 ) f ( 0 ) to find the y -intercept.
• Solve the quadratic equation f ( x ) = 0 f ( x ) = 0 to find the x -intercepts.

Finding the y - and x -Intercepts of a Parabola

Find the y - and x -intercepts of the quadratic f ( x ) = 3 x 2 + 5 x − 2. f ( x ) = 3 x 2 + 5 x − 2.

We find the y -intercept by evaluating f ( 0 ) . f ( 0 ) .

So the y -intercept is at ( 0 , −2 ) . ( 0 , −2 ) .

For the x -intercepts, we find all solutions of f ( x ) = 0. f ( x ) = 0.

In this case, the quadratic can be factored easily, providing the simplest method for solution.

So the x -intercepts are at ( 1 3 , 0 ) ( 1 3 , 0 ) and ( − 2 , 0 ) . ( − 2 , 0 ) .

By graphing the function, we can confirm that the graph crosses the y -axis at ( 0 , −2 ) . ( 0 , −2 ) . We can also confirm that the graph crosses the x -axis at ( 1 3 , 0 ) ( 1 3 , 0 ) and ( −2 , 0 ) . ( −2 , 0 ) . See Figure 14

Rewriting Quadratics in Standard Form

In Example 7 , the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.

Given a quadratic function, find the x - x - intercepts by rewriting in standard form .

• Substitute a a and b b into h = − b 2 a . h = − b 2 a .
• Substitute x = h x = h into the general form of the quadratic function to find k . k .
• Rewrite the quadratic in standard form using h h and k . k .
• Solve for when the output of the function will be zero to find the x - x - intercepts.

Finding the x -Intercepts of a Parabola

Find the x - x - intercepts of the quadratic function f ( x ) = 2 x 2 + 4 x − 4. f ( x ) = 2 x 2 + 4 x − 4.

We begin by solving for when the output will be zero.

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

We know that a = 2. a = 2. Then we solve for h h and k . k .

So now we can rewrite in standard form.

We can now solve for when the output will be zero.

The graph has x -intercepts at ( −1 − 3 , 0 ) ( −1 − 3 , 0 ) and ( −1 + 3 , 0 ) . ( −1 + 3 , 0 ) .

We can check our work by graphing the given function on a graphing utility and observing the x - x - intercepts. See Figure 15 .

We could have achieved the same results using the quadratic formula. Identify a = 2 , b = 4 a = 2 , b = 4 and c = −4. c = −4.

So the x -intercepts occur at ( − 1 − 3 , 0 ) ( − 1 − 3 , 0 ) and ( − 1 + 3 , 0 ) . ( − 1 + 3 , 0 ) .

In a Try It , we found the standard and general form for the function g ( x ) = 13 + x 2 − 6 x . g ( x ) = 13 + x 2 − 6 x . Now find the y - and x -intercepts (if any).

Applying the Vertex and x -Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H ( t ) = − 16 t 2 + 80 t + 40. H ( t ) = − 16 t 2 + 80 t + 40.

• ⓐ When does the ball reach the maximum height?
• ⓑ What is the maximum height of the ball?
• ⓒ When does the ball hit the ground?

The ball reaches a maximum height after 2.5 seconds.

The ball reaches a maximum height of 140 feet.

We use the quadratic formula.

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16 .

Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph.

A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation H ( t ) = −16 t 2 + 96 t + 112. H ( t ) = −16 t 2 + 96 t + 112.

• ⓐ When does the rock reach the maximum height?
• ⓑ What is the maximum height of the rock?
• ⓒ When does the rock hit the ocean?

Access these online resources for additional instruction and practice with quadratic equations.

• Graphing Quadratic Functions in General Form
• Graphing Quadratic Functions in Standard Form
• Quadratic Function Review
• Characteristics of a Quadratic Function

5.1 Section Exercises

Explain the advantage of writing a quadratic function in standard form.

How can the vertex of a parabola be used in solving real-world problems?

Explain why the condition of a ≠ 0 a ≠ 0 is imposed in the definition of the quadratic function.

What is another name for the standard form of a quadratic function?

What two algebraic methods can be used to find the horizontal intercepts of a quadratic function?

For the following exercises, rewrite the quadratic functions in standard form and give the vertex.

f ( x ) = x 2 − 12 x + 32 f ( x ) = x 2 − 12 x + 32

g ( x ) = x 2 + 2 x − 3 g ( x ) = x 2 + 2 x − 3

f ( x ) = x 2 − x f ( x ) = x 2 − x

f ( x ) = x 2 + 5 x − 2 f ( x ) = x 2 + 5 x − 2

h ( x ) = 2 x 2 + 8 x − 10 h ( x ) = 2 x 2 + 8 x − 10

k ( x ) = 3 x 2 − 6 x − 9 k ( x ) = 3 x 2 − 6 x − 9

f ( x ) = 2 x 2 − 6 x f ( x ) = 2 x 2 − 6 x

f ( x ) = 3 x 2 − 5 x − 1 f ( x ) = 3 x 2 − 5 x − 1

For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry.

y ( x ) = 2 x 2 + 10 x + 12 y ( x ) = 2 x 2 + 10 x + 12

f ( x ) = 2 x 2 − 10 x + 4 f ( x ) = 2 x 2 − 10 x + 4

f ( x ) = − x 2 + 4 x + 3 f ( x ) = − x 2 + 4 x + 3

f ( x ) = 4 x 2 + x − 1 f ( x ) = 4 x 2 + x − 1

h ( t ) = −4 t 2 + 6 t − 1 h ( t ) = −4 t 2 + 6 t − 1

f ( x ) = 1 2 x 2 + 3 x + 1 f ( x ) = 1 2 x 2 + 3 x + 1

f ( x ) = − 1 3 x 2 − 2 x + 3 f ( x ) = − 1 3 x 2 − 2 x + 3

For the following exercises, determine the domain and range of the quadratic function.

f ( x ) = ( x − 3 ) 2 + 2 f ( x ) = ( x − 3 ) 2 + 2

f ( x ) = −2 ( x + 3 ) 2 − 6 f ( x ) = −2 ( x + 3 ) 2 − 6

f ( x ) = x 2 + 6 x + 4 f ( x ) = x 2 + 6 x + 4

f ( x ) = 2 x 2 − 4 x + 2 f ( x ) = 2 x 2 − 4 x + 2

For the following exercises, use the vertex ( h , k ) ( h , k ) and a point on the graph ( x , y ) ( x , y ) to find the general form of the equation of the quadratic function.

( h , k ) = ( 2 , 0 ) , ( x , y ) = ( 4 , 4 ) ( h , k ) = ( 2 , 0 ) , ( x , y ) = ( 4 , 4 )

( h , k ) = ( −2 , −1 ) , ( x , y ) = ( −4 , 3 ) ( h , k ) = ( −2 , −1 ) , ( x , y ) = ( −4 , 3 )

( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 2 , 5 ) ( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 2 , 5 )

( h , k ) = ( 2 , 3 ) , ( x , y ) = ( 5 , 12 ) ( h , k ) = ( 2 , 3 ) , ( x , y ) = ( 5 , 12 )

( h , k ) = ( − 5 , 3 ) , ( x , y ) = ( 2 , 9 ) ( h , k ) = ( − 5 , 3 ) , ( x , y ) = ( 2 , 9 )

( h , k ) = ( 3 , 2 ) , ( x , y ) = ( 10 , 1 ) ( h , k ) = ( 3 , 2 ) , ( x , y ) = ( 10 , 1 )

( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 1 , 0 ) ( h , k ) = ( 0 , 1 ) , ( x , y ) = ( 1 , 0 )

( h , k ) = ( 1 , 0 ) , ( x , y ) = ( 0 , 1 ) ( h , k ) = ( 1 , 0 ) , ( x , y ) = ( 0 , 1 )

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts.

f ( x ) = x 2 − 2 x f ( x ) = x 2 − 2 x

f ( x ) = x 2 − 6 x − 1 f ( x ) = x 2 − 6 x − 1

f ( x ) = x 2 − 5 x − 6 f ( x ) = x 2 − 5 x − 6

f ( x ) = x 2 − 7 x + 3 f ( x ) = x 2 − 7 x + 3

f ( x ) = −2 x 2 + 5 x − 8 f ( x ) = −2 x 2 + 5 x − 8

f ( x ) = 4 x 2 − 12 x − 3 f ( x ) = 4 x 2 − 12 x − 3

For the following exercises, write the equation for the graphed quadratic function.

For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function.

For the following exercises, use a calculator to find the answer.

Graph on the same set of axes the functions f ( x ) = x 2 f ( x ) = x 2 , f ( x ) = 2 x 2 f ( x ) = 2 x 2 , and f ( x ) = 1 3 x 2 f ( x ) = 1 3 x 2 .

What appears to be the effect of changing the coefficient?

Graph on the same set of axes f ( x ) = x 2 , f ( x ) = x 2 + 2 f ( x ) = x 2 , f ( x ) = x 2 + 2 and f ( x ) = x 2 , f ( x ) = x 2 + 5 f ( x ) = x 2 , f ( x ) = x 2 + 5 and f ( x ) = x 2 − 3. f ( x ) = x 2 − 3. What appears to be the effect of adding a constant?

Graph on the same set of axes f ( x ) = x 2 , f ( x ) = ( x − 2 ) 2 , f ( x − 3 ) 2 f ( x ) = x 2 , f ( x ) = ( x − 2 ) 2 , f ( x − 3 ) 2 , and f ( x ) = ( x + 4 ) 2 . f ( x ) = ( x + 4 ) 2 .

What appears to be the effect of adding or subtracting those numbers?

The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the function h ( x ) = − 32 ( 80 ) 2 x 2 + x h ( x ) = − 32 ( 80 ) 2 x 2 + x where x x is the horizontal distance traveled and h ( x ) h ( x ) is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally.

A suspension bridge can be modeled by the quadratic function h ( x ) = .0001 x 2 h ( x ) = .0001 x 2 with −2000 ≤ x ≤ 2000 −2000 ≤ x ≤ 2000 where | x | | x | is the number of feet from the center and h ( x ) h ( x ) is height in feet. Use the TRACE feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet.

For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function.

Vertex ( 1 , −2 ) , ( 1 , −2 ) , opens up.

Vertex ( −1 , 2 ) ( −1 , 2 ) opens down.

Vertex ( −5 , 11 ) , ( −5 , 11 ) , opens down.

Vertex ( −100 , 100 ) , ( −100 , 100 ) , opens up.

For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function.

Contains ( 1 , 1 ) ( 1 , 1 ) and has shape of f ( x ) = 2 x 2 . f ( x ) = 2 x 2 . Vertex is on the y - y - axis.

Contains ( −1 , 4 ) ( −1 , 4 ) and has the shape of f ( x ) = 2 x 2 . f ( x ) = 2 x 2 . Vertex is on the y - y - axis.

Contains ( 2 , 3 ) ( 2 , 3 ) and has the shape of f ( x ) = 3 x 2 . f ( x ) = 3 x 2 . Vertex is on the y - y - axis.

Contains ( 1 , −3 ) ( 1 , −3 ) and has the shape of f ( x ) = − x 2 . f ( x ) = − x 2 . Vertex is on the y - y - axis.

Contains ( 4 , 3 ) ( 4 , 3 ) and has the shape of f ( x ) = 5 x 2 . f ( x ) = 5 x 2 . Vertex is on the y - y - axis.

Contains ( 1 , −6 ) ( 1 , −6 ) has the shape of f ( x ) = 3 x 2 . f ( x ) = 3 x 2 . Vertex has x-coordinate of −1. −1.

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• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
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• High School Math Solutions – Quadratic Equations Calculator, Part 2 Solving quadratics by factorizing (link to previous post) usually works just fine. But what if the quadratic equation...

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Module 10: Quadratic Equations and Functions

10.3 – applications of quadratic functions, learning objectives, objects in free fall.

• Determining the width of a border

Finding the maximum and minimum values of a quadratic function

(10.3.1) – solve application problems involving quadratic functions.

Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.

A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation isn’t completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.

A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called initial velocity ) is [latex]−10[/latex] feet per second. (The negative value means it’s heading toward the ground.)

The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after [latex]t[/latex] seconds. About how long does it take for the ball to hit the ground?

When the ball hits the ground, the height is 0. Substitute 0 for [latex]h[/latex].

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is [latex]t[/latex] rather than [latex]x[/latex]. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex]\displaystyle t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex]\large \begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

[latex]t[/latex] is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.

The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

In the next video we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.

Here are some more similar objects in free fall examples.

Example: Applying the Vertex and [latex] x[/latex] -Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+80t+40[/latex].

a. When does the ball reach the maximum height?

b. What is the maximum height of the ball?

c. When does the ball hit the ground?

[latex]\large \begin{array}{c} h=-\frac{80}{2\left(-16\right)} \text{ }=\frac{80}{32}\hfill \\ \text{ }=\frac{5}{2}\hfill \\ \text{ }=2.5\hfill \end{array}[/latex]

The ball reaches a maximum height after 2.5 seconds.

b. To find the maximum height, find the y  coordinate of the vertex of the parabola.

[latex]\large \begin{array}{c}k=H\left(-\frac{b}{2a}\right)\hfill \\ \text{ }=H\left(2.5\right)\hfill \\ \text{ }=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40\hfill \\ \text{ }=140\hfill \end{array}[/latex]

The ball reaches a maximum height of 140 feet.

c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\left(t\right)=0[/latex].

We use the quadratic formula.

[latex]\large \begin{array}{c} t=\frac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)}\hfill \\ \text{ }=\frac{-80\pm \sqrt{8960}}{-32}\hfill \end{array}[/latex]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

[latex]\large \begin{array}{c}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458\hfill & \text{or}\hfill & t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458\hfill \end{array}[/latex]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.

A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+96t+112[/latex].

a. When does the rock reach the maximum height?

b. What is the maximum height of the rock?

c. When does the rock hit the ocean?

a. 3 seconds

b. 256 feet

c. 7 seconds

Applications of quadratic functions: determining the width of a border

The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.

Bob made a quilt that is 4 ft [latex]\times[/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)

Sketch the problem. Since you don’t know the width of the border, you will let the variable [latex]x[/latex] represent the width.

In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.

Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex].

(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)

You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are 10 sq ft of fabric for the border, so set the area of border to be 10.

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

[latex]10=18x+4x^{2}[/latex]

Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}-10\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\large \begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\large \begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

[latex]\displaystyle x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\large \begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative.

The width of the border should be 0.5 ft.

Here is a video which gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Find two numbers [latex]x[/latex] and [latex]y[/latex] whose difference is 100 and whose product is a minimum.

We are trying to find the minimum of the product [latex]P=xy[/latex] of two numbers, such that their difference is 100: [latex]y-x=100[/latex]. First, we rewrite one variable in terms of the other:

[latex]y-x=100 \rightarrow y=100+x[/latex]

Next, we plug in the above relationship between the variables into the first equation:

[latex]P=xy=x(100+x) = 100x+x^2 = x^2+100x[/latex]

As a result, we get a quadratic function [latex]P(x)=x^2+100x[/latex]. The graph of this quadratic function opens upwards, and its vertex is the minimum, So if we find the vertex of this parabola, we will find the minimum product. The vertex is:

[latex]\displaystyle \displaystyle \left(-\frac{b}{2a}, P\left(-\frac{b}{2a}\right)\right) = \left(-\frac{(100)}{2(1)}, P\left(-\frac{(100)}{2(1)}\right)\right) = (-50,-2,500)[/latex]

Thus the minimum of the parabola occurs at [latex]x=-50[/latex], and is [latex]-2,500[/latex]. So one of the numbers is [latex]x=-50[/latex], the other we obtain by plugging in [latex]x=-50[/latex] in to [latex]y-x=100[/latex]:

[latex]\begin{array}{cc}y-(-50)&=&100 \\ y+50 &=& 100 \\ y &=& 50\end{array}[/latex]

[latex]x=-50[/latex] and [latex]y=50[/latex]

Example: Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

• Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[/latex].
• What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as the one above to record the given information. It is also helpful to introduce a temporary variable, W , to represent the width of the garden and the length of the fence section parallel to the backyard fence.

1)  We know we have only 80 feet of fence available, and [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex]. This allows us to represent the width, [latex]W[/latex], in terms of [latex]L[/latex].

[latex]W=80 - 2L[/latex]

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

[latex]\begin{array}{l}\text{ }A&=&LW=L\left(80 - 2L\right)\hfill \\ A\left(L\right)&=&80L - 2{L}^{2}\hfill \end{array}[/latex]

This formula represents the area of the fence in terms of the variable length [latex]L[/latex]. The function, written in general form, is

[latex]A\left(L\right)=-2{L}^{2}+80L[/latex].

2) The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]a[/latex] is the coefficient of the squared term, [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

To find the vertex:

[latex]\large \begin{array}{l}h=-\frac{80}{2\left(-2\right)}\hfill & \hfill & \hfill & \hfill & k=A\left(20\right)\hfill \\ \text{ }=20\hfill & \hfill & \text{and}\hfill & \hfill & \text{ }=80\left(20\right)-2{\left(20\right)}^{2}\hfill \\ \hfill & \hfill & \hfill & \hfill & \text{ }=800\hfill \end{array}[/latex]

The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

Analysis of the Solution

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.

How To: Given an application involving revenue, use a quadratic equation to find the maximum.

• Write a quadratic equation for revenue.
• Find the vertex of the quadratic equation.
• Determine the [latex]y[/latex]-value of the vertex.

Example: Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]p[/latex] for price per subscription and [latex]Q[/latex] for quantity, giving us the equation [latex]\text{Revenue}=pQ[/latex].

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently [latex]p=30[/latex] and [latex]Q=84,000[/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[/latex] and [latex]Q=79,000[/latex]. From this we can find a linear equation relating the two quantities. The slope will be

[latex]\large \begin{array}{c}m=\frac{79,000 - 84,000}{32 - 30}\hfill \\ \text{ }=\frac{-5,000}{2}\hfill \\ \text{ }=-2,500\hfill \end{array}[/latex]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y -intercept.

[latex]\begin{array}{c}\text{ }Q=-2500p+b\hfill & \text{Substitute in the point }Q=84,000\text{ and }p=30\hfill \\ 84,000=-2500\left(30\right)+b\hfill & \text{Solve for }b\hfill \\ \text{ }b=159,000\hfill & \hfill \end{array}[/latex]

This gives us the linear equation [latex]Q=-2,500p+159,000[/latex] relating cost and subscribers. We now return to our revenue equation.

[latex]\begin{array}{c}\text{Revenue}=pQ\hfill \\ \text{Revenue}=p\left(-2,500p+159,000\right)\hfill \\ \text{Revenue}=-2,500{p}^{2}+159,000p\hfill \end{array}[/latex]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

[latex]\large \begin{array}{c}h=-\frac{159,000}{2\left(-2,500\right)}\hfill \\ \text{ }=31.8\hfill \end{array}[/latex]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

[latex]\begin{array}{c}\text{maximum revenue}&=&-2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right)\hfill \\ \text{ }&=&2,528,100\hfill \end{array}[/latex]

This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.

A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?

(credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at [latex]\left(-4,\text{ }7\right)[/latex], so [latex]\left(h\right)x=-\frac{7}{16}{\left(x+4\right)}^{2}+7[/latex]. To make the shot, [latex]h\left(-7.5\right)[/latex] would need to be about 4 but [latex]h\left(-7.5\right)\approx 1.64[/latex]; he doesn’t make it.

• Quadratic Formula Application - Time for an Object to Hit the Ground. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/RcVeuJhcuL0 . License : CC BY: Attribution
• Quadratic Formula Application - Determine the Width of a Border. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/Zxe-SdwutxA . License : CC BY: Attribution
• College Algebra. Authored by : Abramson, Jay, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution . License Terms : Download for free at : http://cnx.org/contents/[email protected]:1/Preface
• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution

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Quadratic Equations

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• Sravanth C.
• Akshay Yadav
• Yash Dev Lamba
• Mehul Arora
• Omkar Kulkarni
• Skanda Prasad
• A Former Brilliant Member

A quadratic equation is a polynomial equation with degree two. In other words, it is an equation of the form \( ax^2 + bx + c =0 \), where \( a \), \(b\) and \(c\) are real numbers and \(a\neq 0\).

Solving by Factoring

Finding quadratic equation from roots, solving by completing the square, solving by quadratic formula, nature of roots of quadratic equation, word problems - basic, biquadratic equations, quadratic equations - problem solving.

Main Article: Factoring Polynomials

We can solve quadratics using factoring and the zero product property . In general, we can rewrite a quadratic as the product of two linear factors such that \( ax^2 + bx + c = a(x+p)(x+q) \). By the zero product property,

\[ \text{if }\ ax^2 + bx + c = a(x+p)(x+q) = 0, \ \text{then either }\ x = -p \ \text{ or }\ x = -q.\]

Now, to factorise a quadratic equation, follow these steps.

\(1)\) We have to break \(b\) (the coefficient of \(x\)) into two terms in such a way that their sum is \(b\) and their product is \(ac:\) \[ax^2+bx+c=0 \implies ax^2+(b_1+b_2)x+c=0 \text{ such that }b_1+b_2=b \text{ and } b_1 \times b_2=ac.\] \(2)\) Next we need to group \(ax^2 + b_1x \text{ and } b_2x+c\) and factorize them in a way that they both have one factor common.

Now we will have the equation transformed into factors. From here on, the solution is easy. We use the zero product property and equate each factor to \(0\), i.e. \((x-\alpha)=0 \implies x=\alpha\) and \((x-\beta)=0 \implies x=\beta\).

Solve \(x^2+5x+6=0\) for \(x\) by the method of factoring. Following the steps mentioned above, we first break the coefficient of \(x\) in two terms such that their sum equals \(5\) and their product equals \(1 \times 6=6\): \[x^2+(2+3)x +6=0,\] where we can observe that \(2+3=5\) and \(2 \times 3=6\): \[\begin{align} x^2+2x+3x+6&=0 \\ x(x+2)+3(x+2)&=0. \end{align}\] Taking out \((x+2)\) as a common factor, we have \[\begin{align} (x+3)(x+2)&=0 \\ x+3&=0 \implies x=-3 \\ x+2&=0 \implies x=-2. \end{align}\] Therefore the two roots of the given equation are \(-3\) and \(-2\). \(_\square\)

Method of Solving a Quadratic Equation by Factorizing: Step 1. Make the given equation free from fractions and radicals and put it into the standard form \(ax^2+bx+c=0.\) Step 2. Factorize \(ax^2+bx+c\) into two linear factors. Step 3. Put each linear factor equal to \(0\) (to apply the zero product rule). Step 4. Solve these linear equations and get two roots of the given quadratic equation.

Solve \(x^2 - x - 6 =0 \) by the method of factoring. We have \[ x^2 - x - 6 = (x-3)(x+2),\] which gives \( x = 3 \) or \( x = -2 \). \(_\square\) Note that the factors of \( x^2 - x - 6 \) are \(1, x^2 - x - 6, x-3,\) and \(x+2\).

Solve the equation \( x^2+3x+2=0 \) for \(x\). We have \[\begin{align} x^2+3x+2 &=0 \\ x^2+2x+x+2 &=0\\ x(x+2)+1(x+2) & =0\\ (x+2)(x+1) & =0. \end{align}\] So, \[\begin{align} (x+2)=0 \text{ or } (x+1)=0 \\ x=-2 \text{ or } x =-1.\ _\square \end{align}\]

Note: We cannot always factor into linear factors using only real numbers . For some quadratics \((\)e.g., \( x^2 + 1 )\), the linear factors require complex numbers:

\[ x^2 + 1 = (x+i)(x-i) .\]

Try the following problems:

No problem found with slug "algebra-2-2"

Find the positive root of the equation \(x^{2}+x-20=0\).

If the value of \( a^2 + 6a -6 \) is \(a\), then find the minimum value of \(a\).

When two places of the variable are given, we have to write them of the form \(\text{(variable - value = 0)}\).

To find the equation from the roots: Step 1. If the variable \(x\) is given, and two values \(x=a\) and \(x=b\) are given, then we have to simplify them to \[x-a=0 \quad\text{ and }\quad x-b=0.\] Step 2. Multiplying the equations and simplifying them, we arrive at this: \[\begin{align} (x-a)(x-b)&=0\\ x^2 -(a+b)x+ab&= 0. \end{align}\]

Find the quadratic equation whose roots are \(2\) and \(-3\) Considering the equation in variable \(x\), we have the following: \[\begin{align} x=2&\implies (x-2)=0\\ x=-3&\implies (x+3)=0. \end{align}\] Multiplying both the equations, we have \[\begin{align} (x-2)(x+3) & =0\\ x(x+3)-2(x+3) & =0\\ x^2+3x-2x-6 & =0\\ x^2+x-6 & =0.\ _\square \end{align}\]

Find the quadratic equation whose roots are \(5\) and \(6\) Considering the equation in variable \(x\), we have the following: \[\begin{align} x&=5 \implies x-5=0\\ x&=6\implies x-6=0. \end{align}\] Multiplying both the equations gives \[\begin{align} (x-5)(x-6) & =0\\ x(x-6)-5(x-6) & =0\\ x^2-6x-5x+30 & =0\\ x^2-11x+30 & =0.\ _\square \end{align}\]

Main Article: Completing The Square

For a quadratic polynomial \(f(x) = ax^2 + bx +c\), completing the square means finding an expression of the form

\[f(x) = a(x-b)^2 + c.\]

Complete the square for the quadratic \( x^2 + 8x + 10 \). Since our middle term is \( 8x \), we know that we will want a perfect square with the form \( (x+4)^2 \), which expands to \( x^2 + 8x + 16 \). Thus, we can do the following: \[\begin{align} x^2 + 8x + 10 &= x^2 + 8x + 16 - 16 + 10\\ &= (x^2 + 8x + 16) - 6 \\ &= ( x+ 4 )^2 -6. \ _\square \end{align} \]

Solve this equation \(2x^{2}+3x+1=0\) by method of completing square. First take \(2\) as common: \(2\left(x^{2}+\frac{3x}{2}\right)+1=0\). Since our middle term is \(\frac 32 x,\) we know that we will want a perfect square with the form \[\left(x+\dfrac{3}{4}\right)^2=x^{2}+\dfrac{3x}{2}+\dfrac{9}{16}.\] So rewrite the whole equation as \[\begin{align} 2\left(x+\dfrac{3x}2 +\dfrac{9}{16}-\dfrac{9}{16}\right)+1 & =0\\ 2\left(x+\dfrac{3x}2 + \left(\dfrac{3}{4}\right)^{2}-\dfrac{9}{16}\right)+1 & =0\\ 2\left(x+\dfrac{3}{4}\right)^{2}-\dfrac{1}{8} & =0\\ \left(x+\dfrac{3}{4}\right)^{2} & =\dfrac{1}{16}. \end{align}\] Thus we have \[\begin{align} x+\dfrac{3}{4} &=\pm \dfrac{1}{4}\\ \Rightarrow x &=-\dfrac{1}{2} \, \text{ or }\, x=-1.\ _\square \end{align}\]

Find the minimum value of \(4x^{2}+8x+16\) for real \(x\).

Main Article: Quadratic Formula

The quadratic formula states that for the equation \( ax^2 + bx + c =0 \), the values of \( x\) are given by the following:

\[ x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}. \]

To see how this formula is derived via completing the square, see Quadratic Formula .

Solve \(5x^2-2x-3=0\) for \(x\). Here, \(a=5, b=-2, c=-3\). Using the quadratic formula, we get \[\begin{align} x &= \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-2) \pm \sqrt{(-2)^2 - 4×5×-3}}{2×5}\\ & = \dfrac {2 \pm \sqrt{4 +60}}{10} = \dfrac {2 \pm \sqrt{64}}{10}\\ & = \dfrac {2 \pm 8}{10}\\ \Rightarrow x & = -0.6 \, \text{ or }\, x=1.\ _\square \end{align}\]

Solve \(x^2-4x+1\) for \(x\). Here, \(a=1, b=-4, c=1\). Using the quadratic formula, we get \[\begin{align} x & = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-4) \pm \sqrt{(-4)^2 - 4×1×1}}{2×1}\\ & = \dfrac {4 \pm \sqrt{16 -4}}{2} =\dfrac {4 \pm \sqrt{12}}{2}\\ &=\dfrac {4 \pm 2×\sqrt {3}}{2}\\ \Rightarrow x & =2+\sqrt3 \, \text{ or }\, x=2-\sqrt3.\ _\square \end{align}\]

Solve \(x^2-20x-69 = 0\) for \(x.\) Substituting the values \(a=1, b=-20, c=-69\) in the quadratic formula, we get \[\begin{align} x & = \dfrac {-(-20) \pm \sqrt {(-20)^2 -4×1×-69}}{2×1}\\ & = \dfrac {20 \pm \sqrt {400 +276}}{2}\\ & = \dfrac {20 \pm \sqrt {676}}{2}\\ & = \dfrac {20 \pm 26}{2}\\ \Rightarrow x & = 23 \, \text{ or }\, x=-3.\ _\square \end{align}\]

\[ x= \dfrac { -b\pm \sqrt { b ^2 - 4ac } }{ 2a } \]

Using the quadratic formula above, find the roots of the equation

\[ x^2−20x−69=0.\]

Solve the quadratic equation

\[ (5x-3)^2 = 32 . \]

Check out the set: 2016 Problems.

Main Article: Parabolas

Here is an example illustrating the above.

Find the equation of a parabola with vertex at \((0,0)\) if its axis of symmetry is the \(y\)-axis and its graph contains the point \(\left(\frac {-1}{2}, 2 \right)\). We write the vertex form of the parabola as \(y = A(x^2)\). Plug in the coordinates of the given point to find \(A\) \[\begin{align} 2 &= A \times\left (\dfrac {-1}{2}\right) ^2 \\ A &= 8\\ \Rightarrow y &= 8x^2.\ _\square \end{align}\]

If the McDonald's logo were stored as a set of pixels, enlargement would quickly result in distorted or pixelated images, which are an eyesore. As such, companies often make vector images of their logos, in which the information is stored as mathematical formulae. Such vector images are easily scaled while maintaining sharp, crisp images.

As a first approximation, the logo is deconstructed and approximated as 2 parabolic curves of the form \( y = -A(x-5)^2 \) and \( y = - A (x+5)^2 \). The McDonald's logo has a height to length ratio of 1.05. What is \(A\)?

As \(x\) ranges over all real values, what is the minimum of

\[ x^2 + ( x + 1) ^2 ? \]

The nature of roots of a quadratic equation can be determined by observing the quadratic formula closely. It basically consists of a discriminant which actually makes the difference in formula and leads us two roots.

We know the quadratic formula is

\[ x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \]

for any quadratic equation written in standard form of \(ax^2+bx+c=0\). The discriminant \(D\) for the quadratic equation is

\[D=b^2-4ac,\]

\[\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}\]

Determine the nature of roots of the following two quadratic equations: \[\begin{align} 2x^2+x-1&=0 \\ x^2-4x+4 &=0. \end{align}\] For the quadratic equation \( 2x^2+x-1=0\): Since \(a=2,b=1,c=-1,\) \[\begin{align} b^2-4ac & =1^2-4 \times 2 \times -1\\ & =9 >0, \end{align}\] which implies that the roots are real and distinct. For the quadratic equation \(x^2-4x+4=0\): Since \(a=1,b=-4,c=4,\) \[\begin{align} b^2-4ac & =(-4)^2-4 \times 1 \times 4\\ & =0, \end{align}\] which implies that the roots are real and repeated. \(_\square\)

Find the value of \(k\) for which the following quadratic polynomial has repeated roots: \[x^2+4x+k.\] We know that if \(D=0,\) then the quadratic polynomial has repeated roots. So, \[\begin{align} b^2-4ac&=0\\ (4)^2-4(1)(k)&=0\\ k&=4.\ _\square \end{align}\]

Show that the equation \(x^2+dx-1=0\) has real and distinct roots for all real values of \(d\). Here, \(a=1, b=d,\) and \(c=-1\). So the discriminant would be \[D=d^2-4×1×-1=d^2+4.\] Since \(d^2\) is a perfect square, it is always greater than or equal to \(0\). So, \[D=d^2+4\geq 4.\] Thus, the discriminant is always greater than \(0\), implying that this equation has distinct real roots for any real value of \(d\). \(_\square\)

Two years ago, a man's age was three times the square of his son's age. In three years, his age will be four times his son's age. Find their present ages. Let the present age of the son be \(x\). Then the son's age two years ago was \(x-2\), and his father's age two years ago was \(3\times (x-2)^2\). This implies the present age of the father is \(\big[3×(x-2)^2\big]+2\), and hence in three years his age will be \(\big[3×(x-2)^2\big] +2+ 3= \big[3×(x-2)^2\big] +5\). The son's age in 3 years will be \(x+3\). According to given conditions, the following holds: \[ \begin{array}{rl} 3(x-2)^2+5 & = 4(x+3) \\ \Rightarrow 3x^2-16x+5 & =0 \\ \Rightarrow (3x-1)(x-5) & =0 \\ \Rightarrow x & =\frac 13, 5. \end{array} \] If \(x\) = \( \frac {1}{3}\), then the son's age 2 years ago would become negative, which is impossible. So, the son's present age is \(x=5\), which implies that the present age of the man is \[\begin{align} 3\times (x-2)^2+2 &= 3\times (5-2)^2+2\\ &= 3\times 3^2+2\\ &= 3×9+2\\ &= 27+2\\ &= 29.\ _\square \end{align}\]

Find two numbers whose sum is \(40\), and product \(375\). Let one number be \(x\). Then, according to the first condition, the second number is \(40-x\). Substituting, the value in the second condition, we get \[\begin{align} x(40-x) &= 375\\ 40x-x^2 &= 375\\ x^2-40x+375 &=0\\ x^2-25x-15x+375 &=0\\ x(x-25)-15(x-25) &=0\\ (x-15)(x-25) &=0\\ \Rightarrow x&=15, x=25. \end{align}\] Therefore, the smaller number is \(15\) and the larger one is \(25\). \(_\square\)

The product of two consecutive positive integers is 90. What is their sum? Since the integers are consecutive, we can rewrite the expression above as \( n(n+1) = 90 \). This gives us the following quadratic equation: \( n^2 +n -90 = 0 \). Factoring, we can see that \[ n^2 +n -90 = (n - 9)(n+10) =0, \] which implies \( n = 9 \). Then the two numbers are 9 and 10, and their sum is 19. \(_\square\)

A teacher, on attempting to arrange the students in the form of a solid square for a mass drill, found that 24 students were left out. When he increased the size of the square by one, he found that he was short of 25 students. Find the number of students.

The difference of the cubes of two consecutive odd positive integers is 400 more than the sum of their squares. Find the sum of the two integers.

Clarification : The odd positive integers are \( 1, 3, 5, 7, 9, \ldots.\) Two consecutive odd positive integers refer to two consecutive numbers in this sequence. It does not refer to two consecutive integers (of which one will not be odd).

Sometimes, the quadratic formula could be useful in solving equations of larger degree.

Solve \( x^{4} - 3x^{2} + 1 = 0 \). That equation isn't something you'd want to factor. So, you could make the substitution \( u = x^{2} \). Then the equation would read \[ u^{2} - 3u + 1 = 0. \] We can solve that with the quadratic formula: \[ u = \dfrac{3 \pm \sqrt{5}}{2}. \] But we're not done yet. We want \( x\), not \(u\). Since \( u = x^{2} = \frac{3 \pm \sqrt{5}}{2} \), solving that equation for \(x\) gives \[x=\pm\sqrt{\dfrac{3 \pm \sqrt{5}}{2}}.\ _\square\]

This section contains miscellaneous problems on quadratic equations for you to try, which will eventually enhance your problem solving skills.

\[ x^2 - 5ax + 100 = \ 0 \\ x^2 - 100x + 5a = \ 0 \]

Find the sum of all possible values of \(x\) that satisfy the equations above. Note that \(a\) is an arbitrary constant.

\[ x(x+1)(x+2)(x+3)=120\]

Find the sum of all the real roots of the equation above.

• Factoring Polynomials
• Completing The Square
• Quadratic Formula

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9.6: Solve Applications of Quadratic Equations

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Learning Objectives

By the end of this section, you will be able to:

• Solve applications modeled by quadratic equations

Be Prepared 9.13

Before you get started, take this readiness quiz.

The sum of two consecutive odd numbers is −100. Find the numbers. If you missed this problem, review Example 2.18.

Be Prepared 9.14

Solve: 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . If you missed this problem, review Example 7.35.

Be Prepared 9.15

Find the length of the hypotenuse of a right triangle with legs 5 inches and 12 inches. If you missed this problem, review Example 2.34.

Solve Applications Modeled by Quadratic Equations

We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.

Let’s first summarize the methods we now have to solve quadratic equations.

Methods to Solve Quadratic Equations

• Square Root Property
• Completing the Square
• Quadratic Formula

As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.

Use a Problem-Solving Strategy.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose a variable to represent that quantity.
• Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Step 5. Solve the equation using algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence

We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one n , then the next one is n + 2. The next one would be n + 2 + 2 or n + 4. This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.

Consecutive even integers Consecutive odd integers 64 , 66 , 68 77 , 79 , 81 n 1 st even integer n 1 st odd integer n + 2 2 nd consecutive even integer n + 2 2 nd consecutive odd integer n + 4 3 rd consecutive even integer n + 4 3 rd consecutive odd integer Consecutive even integers Consecutive odd integers 64 , 66 , 68 77 , 79 , 81 n 1 st even integer n 1 st odd integer n + 2 2 nd consecutive even integer n + 2 2 nd consecutive odd integer n + 4 3 rd consecutive even integer n + 4 3 rd consecutive odd integer

Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.

Example 9.35

The product of two consecutive odd integers is 195. Find the integers.

Try It 9.69

The product of two consecutive odd integers is 99. Find the integers.

Try It 9.70

The product of two consecutive even integers is 168. Find the integers.

We will use the formula for the area of a triangle to solve the next example.

Area of a Triangle

For a triangle with base, b , and height, h , the area, A , is given by the formula A = 1 2 b h . A = 1 2 b h .

Recall that when we solve geometric applications, it is helpful to draw the figure.

Example 9.36

An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.

Try It 9.71

Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches.

Try It 9.72

If a triangle that has an area of 110 square feet has a base that is two feet less than twice the height, what is the length of its base and height?

In the two preceding examples, the number in the radical in the Quadratic Formula was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.

We will use the formula for the area of a rectangle to solve the next example.

Area of a Rectangle

For a rectangle with length, L , and width, W , the area, A , is given by the formula A = LW .

Example 9.37

Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.

Try It 9.73

The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.

Try It 9.74

A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot.?

The Pythagorean Theorem gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.

Pythagorean Theorem

In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, a 2 + b 2 = c 2 .

Example 9.38

Rene is setting up a holiday light display. He wants to make a ‘tree’ in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?

Try It 9.75

The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.

Try It 9.76

The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.

The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, v 0 , propels the object up until gravity causes the object to fall back down.

Projectile motion

The height in feet, h , of an object shot upwards into the air with initial velocity, v 0 v 0 , after t t seconds is given by the formula

h = −16 t 2 + v 0 t h = −16 t 2 + v 0 t

We can use this formula to find how many seconds it will take for a firework to reach a specific height.

Example 9.39

A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.

Try It 9.77

An arrow is shot from the ground into the air at an initial speed of 108 ft/s. Use the formula h = −16 t 2 + v 0 t to determine when the arrow will be 180 feet from the ground. Round the nearest tenth.

Try It 9.78

A man throws a ball into the air with a velocity of 96 ft/s. Use the formula h = −16 t 2 + v 0 t to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

We have solved uniform motion problems using the formula D = rt in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The formula D = rt assumes we know r and t and use them to find D . If we know D and r and need to find t , we would solve the equation for t and get the formula t = D r . t = D r .

Some uniform motion problems are also modeled by quadratic equations.

Example 9.40

Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?

This is a uniform motion situation. A diagram will help us visualize the situation.  

Try It 9.79

MaryAnne just returned from a visit with her grandchildren back east . The trip was 2400 miles from her home and her total time in the airplane for the round trip was 10 hours. If the plane was flying at a rate of 500 miles per hour, what was the speed of the jet stream?

Try It 9.80

Gerry just returned from a cross country trip. The trip was 3000 miles from his home and his total time in the airplane for the round trip was 11 hours. If the plane was flying at a rate of 550 miles per hour, what was the speed of the jet stream?

Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We’ll use a similar scenario now.

Example 9.41

The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Try It 9.81

The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours more than Press #2 to do the job and when both presses are running they can print the job in 4 hours. How long does it take for each press to print the job alone?

Try It 9.82

Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes 3 hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in 2 hours. How long does it take for each hose to fill the hot tub?

Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.

• Word Problems Involving Quadratic Equations
• Quadratic Equation Word Problems
• Applying the Quadratic Formula

Section 9.5 Exercises

Practice makes pefect.

In the following exercises, solve using any method.

The product of two consecutive odd numbers is 255. Find the numbers.

The product of two consecutive even numbers is 360. Find the numbers.

The product of two consecutive even numbers is 624. Find the numbers.

The product of two consecutive odd numbers is 1,023. Find the numbers.

The product of two consecutive odd numbers is 483. Find the numbers.

The product of two consecutive even numbers is 528. Find the numbers.

In the following exercises, solve using any method. Round your answers to the nearest tenth, if needed.

A triangle with area 45 square inches has a height that is two less than four times the base Find the base and height of the triangle.

The base of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the base and height of the triangle.

The area of a triangular flower bed in the park has an area of 120 square feet. The base is 4 feet longer that twice the height. What are the base and height of the triangle?

A triangular banner for the basketball championship hangs in the gym. It has an area of 75 square feet. What is the length of the base and height , if the base is two-thirds of the height?

The length of a rectangular driveway is five feet more than three times the width. The area is 50 square feet. Find the length and width of the driveway.

A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?

A rectangular table for the dining room has a surface area of 24 square feet. The length is two more feet than twice the width of the table. Find the length and width of the table.

The new computer has a surface area of 168 square inches. If the the width is 5.5 inches less that the length, what are the dimensions of the computer?

The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle.

The hypotenuse of a right triangle is 10 cm long. One of the triangle’s legs is three times the length of the other leg. Find the lengths of the two legs of the triangle. Round to the nearest tenth.

A rectangular garden will be divided into two plots by fencing it on the diagonal. The diagonal distance from one corner of the garden to the opposite corner is five yards longer than the width of the garden. The length of the garden is three times the width. Find the length of the diagonal of the garden.

Nautical flags are used to represent letters of the alphabet. The flag for the letter, O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The hypotenuse of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.

Gerry plans to place a 25-foot ladder against the side of his house to clean his gutters. The bottom of the ladder will be 5 feet from the house.How far up the side of the house will the ladder reach?

John has a 10-foot piece of rope that he wants to use to support his 8-foot tree. How far from the base of the tree should he secure the rope?

A firework rocket is shot upward at a rate of 640 ft/sec. Use the projectile formula h = −16 t 2 + v 0 t to determine when the height of the firework rocket will be 1200 feet.

An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula h = −16 t 2 + v 0 t , to determine when height of the arrow will be 400 feet.

A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula h = −16 t 2 + v 0 t + 8 to determine how many seconds it will take for the bullet to hit the ground. (That is, when will h = 0?)

A stone is dropped from a 196-foot platform. Use the formula h = −16 t 2 + v 0 t + 196 to determine how many seconds it will take for the stone to hit the ground. (Since the stone is dropped, v 0 = 0.)

The businessman took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours and each way the trip was 200 miles. What was the speed of the wind that affected the plane which was flying at a speed of 120 mph?

The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 300 miles. If the plane was flying at 125 mph, what was the speed of the wind that affected the plane?

Roy kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?

Rick paddled up the river, spent the night camping, and and then paddled back. He spent 10 hours paddling and the campground was 24 miles away. If Rick kayaked at a speed of 5 mph, what was the speed of the current?

Two painters can paint a room in 2 hours if they work together. The less experienced painter takes 3 hours more than the more experienced painter to finish the job. How long does it take for each painter to paint the room individually?

Two gardeners can do the weekly yard maintenance in 8 minutes if they work together. The older gardener takes 12 minutes more than the younger gardener to finish the job by himself. How long does it take for each gardener to do the weekly yard maintainence individually?

It takes two hours for two machines to manufacture 10,000 parts. If Machine #1 can do the job alone in one hour less than Machine #2 can do the job, how long does it take for each machine to manufacture 10,000 parts alone?

Sully is having a party and wants to fill his swimming pool. If he only uses his hose it takes 2 hours more than if he only uses his neighbor’s hose. If he uses both hoses together, the pool fills in 4 hours. How long does it take for each hose to fill the pool?

Writing Exercises

Make up a problem involving the product of two consecutive odd integers.

ⓐ Start by choosing two consecutive odd integers. What are your integers?

ⓑ What is the product of your integers?

ⓒ Solve the equation n ( n + 2) = p , where p is the product you found in part (b).

ⓓ Did you get the numbers you started with?

Make up a problem involving the product of two consecutive even integers.

ⓐ Start by choosing two consecutive even integers. What are your integers?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Real World Examples of Quadratic Equations

A Quadratic Equation looks like this:

Quadratic equations pop up in many real world situations!

Here we have collected some examples for you, and solve each using different methods:

• Factoring Quadratics
• Completing the Square
• Graphing Quadratic Equations
• The Quadratic Formula
• Online Quadratic Equation Solver

Each example follows three general stages:

• Take the real world description and make some equations
• Use your common sense to interpret the results

Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. when does it hit the ground.

Ignoring air resistance, we can work out its height by adding up these three things: (Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t 2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t 2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t 2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1 :

5t 2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give ac , and add to give b " method in Factoring Quadratics :

ac = −15 , and b = −14 .

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t 2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations , and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a :

• t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

• h = −5t 2 + 14t + 3 = −5(1.4) 2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

• $700,000 for manufacturing set-up costs, advertising, etc
• $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

• at $0, you just give away 70,000 bikes
• at $350, you won't sell any bikes at all
• at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

Profit = −200P 2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square .

Solve: −200P 2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2) 2 = (−460/2) 2 = (−230) 2 = 52900

Step 4 Take the square root on both sides of the equation:

Step 5 Subtract (-230) from both sides (in other words, add 230):

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230 , and you can expect:

• Unit Sales = 70,000 − 200 x 230 = 24,000
• Sales in Dollars = $230 x 24,000 = $5,520,000
• Costs = 700,000 + $110 x 24,000 = $3,340,000
• Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm 2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area of steel after cutting out the 11 × 6 middle:

Let us solve this one graphically !

Here is the graph of 4x 2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm 2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. the river has a current of 2 km an hour. what is the boat's speed and how long was the upstream journey.

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

• Let x = the boat's speed in the water (km/h)
• Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

• when going upstream, v = x−2 (its speed is reduced by 2 km/h)
• when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2) (x+2) :

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x 2 −4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x 2 − 30x − 12 = 0

It is a Quadratic Equation!

Let us solve it using the Quadratic Formula :

Where a , b and c are from the Quadratic Equation in "Standard Form": ax 2 + bx + c = 0

Solve 3x 2 - 30x - 12 = 0

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "R T " is:

1 R T   =   1 R 1 + 1 R 2

In this case, we have R T = 2 and R 2 = R 1 + 3

1 2   =   1 R 1 + 1 R 1 +3

To get rid of the fractions we can multiply all terms by 2R 1 (R 1 + 3) and then simplify:

Yes! A Quadratic Equation!

Let us solve it using our Quadratic Equation Solver .

• Enter 1, −1 and −6
• And you should get the answers −2 and 3

R 1 cannot be negative, so R 1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

And many questions involving time, distance and speed need quadratic equations.

Solving Quadratics Practice Questions

Click here for questions, click here for answers, gcse revision cards.

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