Copyright © Cengage Learning. All rights reserved. Quadratic Equations, Quadratic Functions, and Complex Numbers 9

Copyright © Cengage Learning. All rights reserved. Section 9.4 Graphing Quadratic Functions

3 Objectives Graph a quadratic function of the form f (x) = ax 2 + 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) = ax 2 + bx + c

4 Objectives 5. Solve an application involving a quadratic equation. 6. Solve a quadratic equation using a graphing calculator

5 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) = ax 2 + bx + c (a ≠ 0) is a quadratic function, because its right side is a second-degree polynomial in the variable x. In this section, we will discuss many quadratic functions.

6 Graph a quadratic function of the form f (x) = ax 2 + bx + c using a table of values and identify the vertex 1.

7 Graph a quadratic function of the form f (x) = ax 2 + bx + c using a table of values and identify the vertex A basic quadratic function, is defined by the equation f (x) = x 2. 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

8 Graph a quadratic function of the form f (x) = ax 2 + 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.

9 Example Graph f (x) = x 2 – 3. Compare the graph with Figure 9-3. Figure 9-3

10 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 = x 2 – 3 y = 3 2 – 3 y = 6 Substitute 3 for x.

11 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

12 Example – Solution The resulting parabola is the graph of f (x) = x 2 – 3. The vertex of the parabola is the point V(0, –3). Note that the graph of f (x) = x 2 – 3 looks just like the graph of f (x) = x 2, except that it is 3 units lower. cont’d

13 Graph a quadratic function of the form f (x) = ax 2 + bx + c using a table of values and identify the vertex Graphs of Parabolas The graph of the function f (x) = ax 2 + bx + c (a  0) is a parabola. It opens upward when a > 0, and it opens downward when a < 0.

14 Find the vertex of a parabola by completing the square 2.

15 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) = x 2 – 6x + 8 if we complete the square in the following way. f (x) = x 2 – 6x + 9 – f (x) = (x – 3) 2 – 1 Add 9 to complete the square on x 2 –6x and then subtract 9. Factor x 2 – 6x + 9 and combine like terms.

16 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

17 Find the vertex of a parabola by completing the square f (3) = (3 – 3) 2 – 1 f (3) = 0 2 – 1 f (3) = –1 The vertex of the parabola is the point V(3, –1). The graph appears in Figure 9-8. Substitute 3 for x. Figure 9-8

18 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.

19 Example Find the vertex of the parabola determined by f (x) = 2x 2 + 8x + 2 and graph the parabola. Solution: To find the vertex of the parabola, we will write f (x) = 2x 2 + 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 x 2 equal to 1 by factoring 2 out of the binomial 2x 2 + 8x. Then, we proceed as:

20 Example – Solution f (x) = 2x 2 + 8x + 2 = 2(x 2 + 4x) + 2 = 2(x 2 + 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 2x 2 + 8x. Complete the square on x 2 + 4x. Factor x 2 + 4x + 4. Distribute the multiplication by 2. Simplify and combine like terms.

21 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 cont’d Figure 9-10

22 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

23 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) = ax 2 + bx + c 3.

24 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) = ax 2 + bx + c Much can be determined about the graph of f (x) = ax 2 + bx + c from the coefficients a, b, and c. We summarize these results as follows.

25 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) = ax 2 + bx + c Graphing the Parabola f (x) = ax 2 + 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 ax 2 + bx + c = 0.

26 Example Graph: f (x) = x 2 – 2x – 3. Solution: The equation is in the form f (x) = ax 2 + 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 =.

27 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) = x 2 – 2x – 3 f (1) = 1 2 – 2  1 – 3 = 1 – 2 – 3 = –4 The vertex of the parabola is the point (1, –4). cont’d

28 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) = x 2 – 2x – 3 0 = x 2 – 2x – 3 cont’d

29 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

30 Example – Solution The graph appears in Figure Figure 9-11 cont’d

31 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) = ax 2 + bx + c Comment If the entire parabola is above or below the x-axis, there will be no x-intercepts.

32 Solve an application involving a quadratic equation 4.

33 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.

34 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

35 Example – Solution If the firm manufactures 500 smartphones, the maximum revenue will be cont’d

36 Example – Solution = 50,000 The firm should manufacture 500 smartphones to get a maximum revenue of $50,000. cont’d

37 Solve a quadratic equation using a graphing calculator 5.

38 Solve a quadratic equation using a graphing calculator We can use graphing methods to solve quadratic equations. For example, the solutions of the equation x 2 – x – 3 = 0 are the values of x that will make y = 0 in the quadratic function f (x) = x 2 – 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) = x 2 – x – 3, using a TI84 graphing calculator we will obtain the graph shown in Figure 9-12(a). Figure 9-12(a)

39 Solve a quadratic equation using a graphing calculator We can find the x-intercept exactly (if they are rational) by using the ZERO command found in the CALC menu. Enter values to the left and right of each x-intercept, similar to the steps you followed to find the minimum/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)