Copyright © Cengage Learning. All rights reserved.

Quadratic Equations, Quadratic Functions, and Complex Numbers 9

Copyright © Cengage Learning. All rights reserved.

Section 9.49.4

Graphing Quadratic Functions

3

Objectives

Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex.

1. Find the vertex of a parabola by completing the

square.

Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c.

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22

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4

Objectives

5. Solve an application involving a quadratic equation.

6. Solve a quadratic equation using a graphing calculator.

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55

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Graphing Quadratic Functions

The function defined by the equation f (x) = mx + b is a linear function, because its right side is a first-degree polynomial in the variable x.

The function defined by f (x) = ax2 + bx + c (a ≠ 0) is a quadratic function, because its right side is asecond-degree polynomial in the variable x. In this section, we will discuss many quadratic functions.

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Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex1.

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Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex

A basic quadratic function, is defined by the equationf (x) = x2. Recall that to graph this function, we find several ordered pairs (x, y) that satisfy the equation, plot the pairs, and join the points with a smooth curve. A table of values and the graph appear in Figure 9-3.

Figure 9-3

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Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex

The graph of a quadratic function is called a parabola.

The lowest point (or minimum point) on the parabola that opens upward is called its vertex.

The vertex of the parabola shown in Figure 9-3 is the point V(0, 0).

If a parabola opens downward, its highest point (or maximum point) is the vertex.

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Example

Graph f (x) = x2 – 3. Compare the graph with Figure 9-3.

Figure 9-3

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Example – Solution

To find ordered pairs (x, y) that satisfy the equation, we select several numbers for x and compute the corresponding values of y. Recall that f (x) = y. If we let x = 3, we have

y = x2 – 3

y = 32 – 3

y = 6

Substitute 3 for x.

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Example – Solution

The ordered pair (3, 6) and others satisfying the equation appear in the table shown in Figure 9-4. To graph the function, we plot the points and draw a smooth curve passing through them.

Figure 9-4

cont’d

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Example – Solution

The resulting parabola is the graph of f (x) = x2 – 3. The vertex of the parabola is the point V(0, –3).

Note that the graph of f (x) = x2 – 3 looks just like the graph of f (x) = x2, except that it is 3 units lower.

cont’d

13

Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex

Graphs of Parabolas

The graph of the function f (x) = ax2 + bx + c (a 0) is a parabola. It opens upward when a > 0, and it opens downward when a < 0.

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Find the vertex of a parabola by completing the square

2.

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Find the vertex of a parabola by completing the square

It is easier to graph a parabola when we know the coordinates of its vertex. We can find the coordinates of the vertex of the graph of

f (x) = x2 – 6x + 8

if we complete the square in the following way.

f (x) = x2 – 6x + 9 – 9 + 8

f (x) = (x – 3)2 – 1

Add 9 to complete the square on x2 –6x and then subtract 9.

Factor x2 – 6x + 9 and combine like terms.

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Find the vertex of a parabola by completing the square

Since a > 0 in the original equation, the graph will be a parabola that opens upward. The vertex will be the minimum point on the parabola, and the y-coordinate of the vertex will be the smallest possible value of y.

Because (x – 3)2 0, the smallest value of y occurs when (x – 3)2 = 0 or when x = 3. To find the corresponding value of y, we substitute 3 for x in the equation f (x) = (x – 3)2 – 1 and simplify.

f (x) = (x – 3)2 – 1

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Find the vertex of a parabola by completing the square

f (3) = (3 – 3)2 – 1

f (3) = 02 – 1

f (3) = –1

The vertex of the parabola is thepoint V(3, –1). The graph appearsin Figure 9-8.

Substitute 3 for x.

Figure 9-8

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Find the vertex of a parabola by completing the square

A generalization of this discussion leads to the following fact.

Graphs of Parabolas with Vertex at (h, k)

The graph of an equation of the form

f (x) = a(x – h)2 + k

is a parabola with its vertex at the point with coordinates(h, k). The parabola opens upward if a > 0, and it opens downward if a < 0.

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Example

Find the vertex of the parabola determined by f (x) = 2x2 + 8x + 2 and graph the parabola.

Solution:To find the vertex of the parabola, we will writef (x) = 2x2 + 8x + 2 in the form f (x) = a(x – h)2 + k by completing the square on the right side of the equation.

As a first step, we will make the coefficient of x2 equal to 1 by factoring 2 out of the binomial 2x2 + 8x. Then, we proceed as:

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Example – Solution

f (x) = 2x2 + 8x + 2

= 2(x2 + 4x) + 2

= 2(x2 + 4x + 4 – 4) + 2

= 2[(x + 2)2 – 4] + 2

= 2(x + 2)2 + 2(–4) + 2

= 2(x + 2)2 – 6

Or

f (x) = 2[x–2 (–2)]2 + (–6)

cont’d

Factor 2 out of 2x2 + 8x.

Complete the square on x2 + 4x.

Factor x2 + 4x + 4.

Distribute the multiplication by 2.

Simplify and combine like terms.

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Example – Solution

Since h = –2 and k = –6, the vertex of the parabola is the point V(–2, –6). Since a = 2, the parabola opens upward. In this case, the vertex will be the minimum point on the graph.

We can select numbers on either side of x = –2 to construct the table shown in Figure 9-10.

cont’d

Figure 9-10

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Example – Solution

To find the y-intercept, we substitute 0 for x in our original equation and solve for y: When x = 0, y = 2. Thus, the y-intercept is (0, 2).

We determine more ordered pairs, plot the points, and draw the parabola.

cont’d

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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c

3.

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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c

Much can be determined about the graph off (x) = ax2 + bx + c from the coefficients a, b, and c. We summarize these results as follows.

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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c

Graphing the Parabola f (x) = ax2 + bx + c

1. If a > 0, the parabola opens upward and the vertex is the minimum. If a < 0, the parabola opens downward and the vertex is the maximum.

2. The coordinates of the vertex are

3. The axis of symmetry is the vertical line

4. The y-intercept is (0, c).

5. The x-intercepts (if any) are determined by the solutions of ax2 + bx + c = 0.

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Example

Graph: f (x) = x2 – 2x – 3.

Solution:

The equation is in the form f (x) = ax2 + bx + c, with a = 1, b = –2, and c = –3.

Since a > 0, the parabola opens upward. To find the x-coordinate of the vertex, we substitute the values for a and b into the formula x = .

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Example – Solution

The x-coordinate of the vertex is x = 1. This is also the equation for the axis of symmetry. To find the y-coordinate,

we can find = f (1) by substituting 1 for x in the

equation and solving for y.

f (x) = x2 – 2x – 3

f (1) = 12 – 2 1 – 3

= 1 – 2 – 3

= –4

The vertex of the parabola is the point (1, –4).

cont’d

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Example – Solution

To graph the parabola, identify several other points with coordinates that satisfy the equation.

One easy point to find is the y-intercept. It is the value of y when x = 0. Thus, the parabola passes through the point (0, –3).

To find the x-intercepts of the graph, we set f (x) equal to 0 and solve the resulting quadratic equation:

f (x) = x2 – 2x – 3

0 = x2 – 2x – 3

cont’d

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Example – Solution

0 = (x – 3)(x + 1)

x – 3 = 0 or x + 1 = 0

x = 3 x = –1

Since the x-intercepts of the graph are (3, 0) and (–1, 0), the graph passes through these points.

Factor.

Set each factor equal to 0.

cont’d

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Example – Solution

The graph appears in Figure 9-11.

Figure 9-11

cont’d

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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c

Comment

If the entire parabola is above or below the x-axis, there will be no x-intercepts.

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Solve an application involving a quadratic equation4.

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Example – Finding Maximum Revenue

An electronics firm manufactures a high-quality smartphone. Over the past 10 years, the firm has learned

that it can sell x smartphones at a price of

dollars. How many smartphones should the firm manufacture and sell to maximize its revenue? Find the maximum revenue.

Solution:

The revenue obtained is the product of the number of smartphones that the firm sells (x) and the price of each

smartphone .

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Example – Solution

Thus, the revenue R is given by the formula

Since the graph of this function is a parabola that opens downward, the maximum value of R will be the value of R determined by the vertex of the parabola. Because the

x-coordinate of the vertex is at x = , we have

cont’d

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Example – Solution

If the firm manufactures 500 smartphones, the maximum revenue will be

cont’d

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Example – Solution

= 50,000

The firm should manufacture 500 smartphones to get a maximum revenue of $50,000.

cont’d

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Solve a quadratic equation using a graphing calculator

5.

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Solve a quadratic equation using a graphing calculator

We can use graphing methods to solve quadratic equations. For example, the solutions of the equation x2 – x – 3 = 0 are the values of x that will make y = 0 in the quadratic function f (x) = x2 – x – 3. To approximate these values, we graph the quadratic function and identify the x-intercepts.

If we use window values of x = [–10, 10] and y = [–10, 10] and graph the function f (x) = x2 – x – 3, using a TI84 graphing calculator wewill obtain the graph shown inFigure 9-12(a). Figure 9-12(a)

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Solve a quadratic equation using a graphing calculator

We can find the x-intercept exactly(if they are rational) by using theZERO command found in the CALCmenu. Enter values to the left andright of each x-intercept, similar to thesteps you followed to find theminimum/maximum.

In this case the x-intercepts are irrational. To find the exact values, we would have to use the quadratic formula.

Figure 9-12(b)