3.1

1. 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

2. Finding the Zeros of a Quadratic Function Examples 1 and 2 and more Find the zeros of f(x) = x2 − x − 6. Solution by Factoring f(x) = x2 − x − 6 = (x – 3)(x + 2) = 0 has solutions x = 3 and x = − 2 Solution Using the Quadratic Formula Solution by Graphing Notice that the graph crosses the horizontal axis at x = 3 and x = − 2 Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

3. Finding the Zeros of a Quadratic Function Example 3 The figure shows a graph of y = h(x) = − ½ x2 − 2. What happens if we try to use algebra to find the zeros of h? Solution We solve the equation h(x) = − ½ x2 − 2 = 0. So − ½ x2 = 2 x2 = −4 x = ± Since is not a real number, there are no real solutions, so h has no real zeros. This corresponds to the fact that the graph of h in the figure does not cross the x-axis. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

4. Concavity and Quadratic Functions Example 3 Let f(x) = x2. Find the average rate of change of f over the intervals of length 2 starting at x = −4 and ending at x = 4. What do these rates tell you about the concavity of the graph of f? Solution Between x = −4 and x = −2, we have average rate of change of f = (f(-2) – f(-4))/((-2) – (-4)) = – 6. Between x = −2 and x = 0, we have average rate of change of f = (f(0) – f(-2))/(0 – (-2)) = – 2. Between x = 0 and x = 2, we have average rate of change of f = (f(2) – f(0))/(2– 0) = 2. Between x = 2 and x = 4, we have average rate of change of f = (f(4) – f(2))/(4 – 2) = 6. Since the rates of change are increasing, the graph is concave up. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

5. Concavity and Quadratic Functions Example 3 continued Graph of f(x) = x2showing the average rate of change of f over the intervals of length 2 starting at x = −4 and ending at x = 4. Solution y = f(x) = x2 Slope = -6 Slope = 6 Since the rates of change are increasing, the graph is concave up. Slope = 2 Slope = -2 Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

6. Finding a Formula From the Zeros and Vertical Intercept y = f(x) Example 3 Find the equation of the parabola in the figure using the factored form. Solution Since the parabola has x-intercepts at x = 1 and x = 3, its formula can be written as f(x) = a(x − 1)(x − 3). Substituting x = 0, y = 6 gives 6 = a(3), resulting in a = 2. Thus, the formula is f(x) = 2(x − 1)(x − 3). Multiplying out gives f(x) = 2x2 − 8x + 6. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

7. Formulas for Quadratic Functions The graph of a quadratic function is a parabola. • The standard form for a quadratic function is y = ax2 + bx + c, where a, b, c are constants, a≠ 0. The parabola opens upward if a > 0 or downward if a < 0, and it intersects the y-axis at c. • The factored form, when it exists, is y = a(x − r)(x − s), where a, r, s are constants, a≠ 0. The parabola intersects the x-axis at x = r and x = s. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

8. 3.2 THE VERTEX OF A PARABOLA Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

9. The Vertex Form of a Quadratic Function Example 1 (a) Sketch f(x) = (x + 3)2 − 4, and indicate the vertex. (b) Estimate the coordinates of the vertex from the graph. (c) Explain how the formula for f can be used to obtain the minimum of f. Solution (a) (b) The vertex of f appears to be about at the point (−3,−4). y = f(x) (c) Note that (x + 3)2 is always positive or zero, so (x + 3)2 takes on its smallest value when x+ 3 = 0, that is, at x = −3. At this point (x + 3)2 − 4 takes on its smallest value, f(−3) = (−3 + 3)2 − 4 = 0 − 4 = −4. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

10. The Vertex Form of a Quadratic Function The vertex form of a quadratic function is y = a(x − h)2 + k, where a, h, k are constants, a≠ 0. The graph of this quadratic function has vertex (h, k) and axis of symmetry x = h. To convert from vertex form to standard form, we multiply out the squared term. To convert from standard form to vertex form, we complete the square. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

11. The Vertex Form of a Quadratic Function Example 2 (a) Put the quadratic function into vertex form by completing the square and then graph it: (a) s(x) = x2 − 6x + 8 Solution (a) To complete the square, find the square of half of the coefficient of the x-term, (−6/2)2= 9. Add and subtract this number after the x-term: s(x) = x2 − 6x + 8 = x2 − 6x +( 9 – 9) + 8 = (x2 − 6x + 9) – 9 + 8 = (x – 3)2 − 1 The vertex of s is (3,−1) and the axis of symmetry is the vertical line x = 3. The parabola opens upward. x = 3 y = s(x) (3, -1) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

12. Modeling with Quadratic Functions Example 4 For t in seconds, the height of a baseball in feet is given by the formula y = f(t) = −16t2 + 47t + 3. Using algebra, find the maximum height reached by the baseball and the time that height is reached. Solution The maximum height is at the vertex, so we complete the square to write the function in vertex form: y = f(t) = −16(t2 + 47/16 t + 3/16) = −16(t2 + 47/16 t + (-47/32)2–(-47/32)2 + 3/16) = −16((t – 47/32)2 −2401/1024) = −16(t – 47/32)2 − 2401/64 Thus, the vertex is at the point ( 47/32 , 2401/64 ), or approximately (1.469, 37.516). This means that the ball reaches its maximum height of 37.516 feet at t = 1.469 seconds. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally