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3.1

3.1. INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS. The baseball height function is an example of a quadratic function , w hose general form is . The graph of a quadratic function is a parabola. Finding the Zeros of a Quadratic Function.

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3.1

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  1. 3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS

  2. The baseball height function is an example of a quadratic function, whose general form is . The graph of a quadratic function is a parabola.

  3. 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

  4. 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.

  5. 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.

  6. 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

  7. 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(-1)(-3) = 3a, resulting in a = 2. Thus, the formula is f(x) = 2(x − 1)(x − 3). Multiplying out gives f(x) = 2x2 − 8x + 6.

  8. 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.

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