How do I find a quadratic function

Question

Find a quadratic function whose x-intercepts are -5 and 3 with a=1; a=2; a=-2; a=5 and how does the value of a affect the intercepts, the axis of symmetry and the vertex? Compare the x-coordinate of the vertex with the midpoint of the x-intercepts. What might you conclude?

Answer 1

y=a(x+5)(x-3)=a(x²+2x-15). This can also be written y=a(x²+2x+1-16)=a((x+1)²-16).

So y+16a=a(x+1)², which has its vertex at (-1,-16a). The axis of symmetry is x=-1.

The x-intercepts are -5 and 3.

When a changes it doesn't affect the x-intercepts or the axis of symmetry because they are independent of a. But it does affect the y-coordinate of the vertex = -16a. When a goes from positive to negative the parabola flips over the x-axis so the vertex becomes the highest point, whereas when a is positive, the vertex is the lowest point. The axis of symmetry lies midway between the x intercepts, because of the symmetry. The midpoint is the average of the x-intercepts=½(3+(-5))=½(-2)=-1, making x=-1 the axis of symmetry. The focus and vertex always lie on the axis of symmetry, which means that their x-coordinate in this case is -1.

For the given values of a=1, 2, -2, 5 the vertex changes respectively to:

(-1,-16), (-1,-32), (-1,32), (-1,-80).