Context: determine the equation of a quadratic function with x intercepts of -2 and 4, that passes throught the point E(2,-16). the quadratic function needs to be in the form y=a(x-h)^2+k Answer question

To find the equation of a quadratic function in the form y = a(x-h)² + k, we first need to find the values of h, k, and a.

1. The x-intercepts are -2 and 4. The vertex form of a parabola is symmetric, so the h value (which is the x-coordinate of the vertex) is halfway between the x-intercepts. We can find this by averaging the x-intercepts: h = (-2 + 4) / 2 = 1.

2. We know that the parabola passes through the point E(2, -16). Since the x-coordinate of this point is not the same as h, we can't immediately find k. However, we can use this point to find a.

3. Substituting x = 2, y = -16, and h = 1 into the equation y = a(x-h)² + k gives us -16 = a(2-1)² + k. Simplifying this gives -16 = a + k.

4. To find k, we need another point on the parabola. We know that the parabola crosses the y-axis when x = 0. Since the x-intercepts are -2 and 4, the y-coordinate at x = 0 is 0 (because the x-intercepts are the points where the parabola crosses the x-axis, i.e., where y = 0). So, we have another point on the parabola: (0, 0).

5. Substituting x = 0, y = 0, and h = 1 into the equation y = a(x-h)² + k gives us 0 = a(0-1)² + k. Simplifying this gives 0 = a + k.

6. We now have a system of two equations: -16 = a + k and 0 = a + k. Solving this system gives a = -16 and k = 0.

7. So, the equation of the quadratic function is y = -16(x-1)² + 0, or y = -16(x-1)².