1 Copyright © Cengage Learning. All rights reserved. 3 Functions and Graphs 3.6 Quadratic Functions

2 Quadratic Functions If a  0, then the graph of y = ax 2 is a parabola with vertex at the origin (0,0), a vertical axis, opening upward if a > 0 or downward if a < 0. In this section we show that the graph of an equation of the form y = ax 2 + bx + c can be obtained by vertical and/or horizontal shifts of the graph of y = ax 2 and hence is also a parabola.

3 Quadratic Functions An important application of such equations is to describe the trajectory, or path, of an object near the surface of the earth when the only force acting on the object is gravitational attraction.

4 Quadratic Functions If b = c = 0 in the preceding definition, then f (x) = ax 2, and the graph is a parabola with vertex at the origin. If b = 0 and c ≠ 0, then f (x) = ax 2 + c

5 Example 1 – Sketching the graph of a quadratic function Sketch the graph of f if (a) f (x) = x 2 (b) f (x) = x Solution: (a) Since f is even, the graph of f (that is, of y = x 2 ) is symmetric with respect to the y-axis. It is similar in shape to but wider than the parabola y = –x 2.Several points on the graph are (0,0),(1, ), (2, –2), and (3, ).

6 Example 1 – Solution Plotting and using symmetry, we obtain the sketch in Figure 2. cont’d Figure 2

7 Example 1 – Solution (b) To find the graph of y = x 2 + 4, we shift the graph of y = x 2 upward a distance 4, obtaining the sketch in Figure 3. cont’d Figure 3

8 Quadratic Functions If f (x) = ax 2 + bx + c and b  0, then, by completing the square, we can change the form to f (x) = a (x – h) 2 + k for some real numbers h and k. This technique is illustrated in the next example.

9 Example 2 – Expressing a quadratic function as f (x) = a(x – h) 2 + k If f (x) = 3x x + 50, express f (x) in the form a (x – h) 2 + k. Solution 1: Before completing the square, it is essential that we factor out the coefficient of x 2 from the first two terms of f (x), as follows: f (x) = 3x x + 50 = 3(x 2 + 8x + ) + 50 given factor out 3 from 3x x

10 Example 2 – Solution 1 We now complete the square for the expression x 2 + 8x within the parentheses by adding the square of half the coefficient of x—that is, or 16. However, if we add 16 to the expression within parentheses, then, because of the factor 3, we are actually adding 48 to f (x). Hence, we must compensate by subtracting 48: f (x) = 3(x 2 + 8x + ) + 50 cont’d given

11 Example 2 – Solution 1 = 3(x 2 + 8x + 16) + (50 – 48) = 3(x + 4) The last expression has the form a (x – h) 2 + k with a = 3, h = –4, and k = 2. cont’d complete the square for x 2 + 8x equivalent equation

12 Example 2 – Solution 2 We begin by dividing both sides by the coefficient of x 2. f (x) = 3x x + 50 cont’d given divide by 3 add and subtract 16, the number that completes the square for x 2 + 8x

13 Example 2 – Solution 2 = (x + 4) 2 + f (x) = 3(x + 4) cont’d equivalent equation multiply by 3

14 Quadratic Functions If f (x) = ax 2 + bx + c, then, by completing the square as in Example 2, we see that the graph of f is the same as the graph of an equation of the form y = a (x – h) 2 + k. The graph of this equation can be obtained from the graph of y = ax 2 shown in Figure 4(a) by means of a horizontal and a vertical shift. Figure 4(a)

15 Quadratic Functions First, as in Figure 4(b), we obtain the graph of y = a (x – h) 2 by shifting the graph of y = ax 2 either to the left or to the right, depending on the sign of h (the figure illustrates the case with h > 0). Figure 4(b)

16 Quadratic Functions Next, as in Figure 4(c), we shift the graph in (b) vertically a distance | k | (the figure illustrates the case with k > 0). Figure 4(c)

17 Quadratic Functions It follows that the graph of a quadratic function is a parabola with a vertical axis. The sketch in Figure 4(c) illustrates one possible graph of the equation y = ax 2 + bx + c. If a > 0, the point (h, k) is the lowest point on the parabola, and the function f has a minimum value f (h) = k. If a < 0, the parabola opens downward, and the point (h, k) is the highest point on the parabola. In this case, the function f has a maximum value f (h) = k.

18 Quadratic Functions We have obtained the following result. For convenience, we often refer to the parabola y = ax 2 + bx + c when considering the graph of this equation.

19 Example 3 – Finding a standard equation of a parabola Express y = 2x 2 – 6x + 4 as a standard equation of a parabola with a vertical axis. Find the vertex and sketch the graph. Solution: y = 2x 2 – 6x + 4 = 2(x 2 – 3x + ) + 4 = given factor out 2 from 2x 2 – 6x complete the square for x 2 – 3x

20 Example 3 – Solution = The last equation has the form of the standard equation of a parabola with a = 2, h = and k = Hence, the vertex V (h, k) of the parabola is Since a = 2 > 0, the parabola opens upward. To find the y-intercept of the graph of y = 2x 2 – 6x + 4, we let x = 0, obtaining y = 4. cont’d equivalent equation

21 Example 3 – Solution To find the x-intercepts, we let y = 0 and solve the equation 2x 2 – 6x + 4 = 0 or the equivalent equation 2(x – 1)(x – 2) = 0, obtaining x = 1 and x = 2. Plotting the vertex and using the x- and y-intercepts provides enough points for a reasonably accurate sketch (see Figure 5). cont’d Figure 5

22 Quadratic Functions If a parabola y = ax 2 + bx + c has x-intercepts x 1 and x 2, as illustrated in Figure 7 for the case a < 0, then the axis of the parabola is the vertical linex = (x 1 + x 2 )/2 through the midpoint of (x 1, 0) and (x 2, 0). Figure 7

23 Quadratic Functions Therefore, the x-coordinate h of the vertex (h, k) is h = (x 1 + x 2 )/2. Some special cases are illustrated in Figures 5 and 6. In the following example we find an equation of a parabola from given data. Figure 6

24 Example 5 – Finding an equation of a parabola with a given vertex Find an equation of a parabola that has vertex V(2,3) and a vertical axis and passes through the point (5,1). Solution: Figure 8 shows the vertex V, the point (5, 1), and a possible position of the parabola. Using the standard equation y = a (x – h) 2 + k Figure 8

25 Example 5 – Solution with h = 2 and k = 3 gives us y = a (x – 2) To find a, we use the fact that (5, 1) is on the parabola and so is a solution of the last equation. Thus, 1 = a(5 – 2) 2 + 3, or Hence, an equation for the parabola is cont’d

26 Quadratic Functions The next theorem gives us a simple formula for locating the vertex of a parabola. It is unnecessary to remember the formula for the y-coordinate of the vertex of the parabola in the preceding result.

27 Quadratic Functions Once the x-coordinate has been found, we can calculate the y-coordinate by substituting –b/(2a) for x in the equation of the parabola.

28 Example 6 – Finding the vertex of a parabola Find the vertex of the parabola y = 2x 2 – 6x + 4. Solution: We considered this parabola in Example 3 and found the vertex by completing the square. We shall use the vertex formula with a = 2 and b = –6, obtaining the x-coordinate = = =

29 Example 6 – Solution We next find the y-coordinate by substituting for x in the given equation: Thus, the vertex is (see Figure 5). cont’d Figure 5

30 Quadratic Functions Since the graph of f (x) = ax 2 + bx + c for a ≠ 0 is a parabola, we can use the vertex formula to help find the maximum or minimum value of a quadratic function. Specifically, since the x-coordinate of the vertex V is –b/(2a), the y-coordinate of V is the function value f (–b/(2a)). Moreover, since the parabola opens downward if a 0, this function value is the maximum or minimum value, respectively, of f.

31 Quadratic Functions We may summarize these facts as follows.

32 Example 8 – Finding the maximum value of a quadratic function A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides so that they are perpendicular to the sheet. How many inches should be turned up to give the gutter its greatest capacity? Solution: The gutter is illustrated in Figure 9. Figure 9

33 Example 8 – Solution If x denotes the number of inches turned up on each side, the width of the base of the gutter is 12 – 2x inches. The capacity will be greatest when the cross-sectional area of the rectangle with sides of lengths x and 12 – 2x has its greatest value. Letting f (x) denote this area, we have f (x) = x(12 – 2x) = 12x – 2x 2 cont’d

34 Example 8 – Solution = – 2x x, which has the form f (x) = ax 2 + bx + c with a = –2, b = 12, and c = 0. Since f is a quadratic function and a = –2 < 0, it follows from the preceding theorem that the maximum value of f occurs at x = = = 3. cont’d

35 Example 8 – Solution Thus, 3 inches should be turned up on each side to achieve maximum capacity. As an alternative solution, we may note that the graph of the function f (x) = x(12 – 2x) has x-intercepts at x = 0 and x = 6. Hence, the average of the intercepts, is the x-coordinate of the vertex of the parabola and the value that yields the maximum capacity. cont’d

36 Quadratic Functions When working with quadratic functions, we are often most interested in finding the vertex and the x-intercepts. Typically, a given quadratic function closely resembles one of the three forms listed in the following chart. Relationship Between Quadratic Function Forms and Their Vertex and x-intercepts

37 Quadratic Functions If the radicands in (1) or (3) are negative, then there are no x-intercepts. To find the x-intercepts with form (1), use the special quadratic equation. If you have a quadratic function in form (3) and want to find the vertex and the x-intercepts, it may be best to first find the x-intercepts by using the quadratic formula.

38 Quadratic Functions Then you can easily obtain the x-coordinate of the vertex, h, since Of course, if the function in form (3) is easily factorable, it is not necessary to use the quadratic formula.