Functions and Their Graphs: Quadratic Functions and Their Applications

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2. Functions and Their Graphs Chapter 12

3. 12 Functions and Their Graphs 12.1 Relations and Functions 12.2 Graphs of Functions and Transformations 12.3 Quadratic Functions and Their Graphs 12.4 Applications of Quadratic Functions and Graphing Other Parabolas. 12.5 The Algebra of Functions 12.6 Variation

4. 12.4 Applications of Quadratic Functions and Graphing Other Parabolas. If the parabola opens upward, the vertex is the lowest point on the parabola. If the parabola opens downward, the vertex is the highest point on the parabola.

5. Example 1 Let f (x)= ‒ x2 + 4x + 2. a) Does the function attain a minimum or maximum value at its vertex? Since a = ‒ 1, the graph of f(x) will open downward. Therefore, the vertex will be the highest point on the parabola. The function will attain its maximum value at the vertex. b). Find the vertex of the graph of f (x). Use to find the x-coordinate of the vertex. For f (x)= ‒ x2+ 4x+ 2 The y-coordinate of the vertex is f (2). f (2)= ‒ (2)2+ 4(2)+ 2 = ‒ 4+ 8 + 2=6 • The vertex is (2,6)

6. Example 1-continued c) What is the minimum or maximum value of the function? f (x) has no minimum value. The maximum value of the function is 6, the y-coordinate of the vertex. (The largest y-value of the function is 6.) We say that the maximum value of the function is 6 and that it occurs at x = 2 (the x-coordinate of the vertex). d) Graph the function to verify parts a)-c). From the graph of f (x), we can see that our conclusions in parts a)- c) make sense.

7. Given a Quadratic Function, Solve an Applied Problem Involving A Maximum o Minimum Value. Example 2 A ball is thrown upward from a height of 24 ft. The height h of the ball (in feet) t sec after the ball is released is given by a) How long does it take the ball to reach its maximum height? b) What is the maximum height attained by the ball? Solution a). To determine how long it takes the ball to reach its maximum height, we . must find the t-coordinate of the vertex. The ball will reach its maximum height after sec.

8. Example 2-continued Solution b). The maximum height the ball reaches is the y-coordinate (or h(t)-coordinate) ofthe vertex. Since the ball attains its maximum height when , find • The ball reaches a maximum height of 28 ft.

9. Graph Parabolas of the Form x = a( y ‒k)2 + h Not all parabolas are functions. Parabolas can open in the x-direction as illustrated below. Parabolas that open in the y-direction, or vertically, result from the functions If we interchange the x and y, we obtain the equations The graphs of these equations are parabolas that open in the x-direction, or horizontally.

10. Example 3a Solution 1) h = ‒ 1and k = ‒ 2. The vertex is ( ‒1, ‒ 2). 2) The axis of symmetry is y = ‒ 2. 3). a = +1, so the parabola opens to the right. It is the same width as To find the x-intercept, let y = 0 and solve for x. x = (y + 2)2 ‒ 1 x = (0 + 2)2 ‒ 1 • The x-intercept is (3,0) x = 4 ‒ 1= 3

11. Example 3a-continued Solution Find the y-intercepts by substituting 0 for x and solving for y. x = (y + 2)2 ‒ 1 0 = (y + 2)2 ‒ 1 Substitute 0 for x 1 = (y + 2)2 Add 1 ± 1 = y + 2 Square Root Property 1 = y + 2 or ‒1 = y ‒ 1 = y + 2 ‒ 3 = y The y-intercepts are (0,‒ 3)and (0,‒ 1) Use the axis of symmetry to locate the point (3,‒ 4) on the graph.

12. Example 3b Solution 1) h = 4and k = 2. The vertex is ( 4, 2). 2) The axis of symmetry is y = 2. 3). a = ‒ 2, so the parabola opens to the left. It is narrower than To find the x-intercept, let y = 0 and solve for x. x = ‒ 2 (y ‒ 2)2 + 4 x = ‒ 2 (0 ‒ 2)2 + 4 • The x-intercept is (‒4,0) x = ‒2(4) + 4 = ‒ 4

13. Example 3b-continued Solution Find the y-intercepts by substituting 0 for x and solving for y. x = ‒ 2 (y ‒ 2)2 + 4 0 = ‒ 2 (y ‒ 2)2 + 4 Substitute 0 for x ‒ 4 = ‒ 2 (y ‒ 2)2 Subtract 4 2 = (y ‒ 2)2 Divide by ‒2 Square Root Property Add 2 The y-intercepts are Use the axis of symmetry to locate the point (‒4, 4) on the graph.

14. Rewrite x = ay2 + by + c as x = a( y ‒k)2 + h by Completing the Square. Example 4 Rewrite x = 2y2 ‒ 4y + 8 in the form x = a( y ‒k)2 + h by completing the square. Solution To complete the square, follow the same procedure used for quadratic functions. Step 1: Divide the equation by 2 so that the coefficient of y2 is 1. Step 2: Separate the constant from the variable terms using parentheses.

15. Example 4-continued Step 3: Complete the square for the quantity in parentheses. Add 1 inside the parentheses and subtract 1 from the 4. Step 4: Factor the expression inside the parentheses. Step 5: Solve the equation for x by multiplying by 2.

16. Find the Vertex of the Graph x= ay2 + by + c Using and Graph the Equation Example 5 Graph x = y2 ‒ 2y + 5. Find the vertex using the vertex formula. Find the x- and y-intercepts. Solution Since this equation is solved for x and is quadratic in y, it opens in the x-direction. a = 1 so it opens to the right. Use the vertex formula to find the y-coordinate of the vertex. Substitute y = 1 into x = y2 ‒ 2y + 5 to find the x-coordinate of the vertex x= (1)2 ‒ 2(1) + 5 x= 1 ‒ 2 + 5 = 4 The vertex is (4,1). Since the vertex is (4, 1) and the parabola opens to the right, the graph has no y-intercepts.

17. Example 5-continued To find the x-intercept, lety=0and solve for x. x = y2 ‒ 2y + 5 • The x-intercept is (5, 0). x = 02 ‒ 2(0) + 5 x = 5 Find another point on the parabola by choosing a value for y that is close to the y-coordinate of the vertex. Let y= ‒1. Find x x = (‒1)2 ‒ 2(‒1) + 5 x = 1 + 2 + 5 = 8 Another point on the parabola is (8, 1). Use the axis of symmetry to locate the additional points (5, 2) and (8, 3).