Question

How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (0.5,0) & (3.5,0)?

Answer 1

Express in vertex form as `f(x) = a(x-2)^2+9`
and solve for `a` to find `a = -4`, so

`f(x) = -4(x-2)^2+9 = -4x^2+16x-7`

Explanation:

In vertex form, the function takes the form:

`f(x) = a(x-2)^2+9` for some constant `a`

`0 = f(0.5) = a(0.5-2)^2+9 = 2.25a+9`

So `2.25a = -9` and `a = -9/2.25 = -4`

So

`f(x) = -4(x-2)^2+9`

`=-4(x^2-4x+4)+9`

`=-4x^2+16x-16+9`

`=-4x^2+16x-7`