What is the quadratic function that is created with roots -3 and 1 and a vertex at (-1, -8)?

Question

anonymous · Answer

in root form (if there is such a thing), you'd write it like this...
$$f(x)=a(x-r_1)(x-r_2)$$where $r_1 \text{ & } r_2$ are the roots and $\text{a}$ is a constant.

to find the value of the constant, simply plug in the values for the roots and the given point, in this case the vertex, and solve for a.

anonymous · Answer

So  f(x) = a(-1+3)(-1+8)
f(x) = a(2)(7)
f(x) = 14a
Where do I go from there?

anonymous · Answer

$$f(-1) = y = -8$$
so it should be like this...$$f(-1)=a(-1+3)(-1-1) \Rightarrow -8 = a(2)(-2) \Rightarrow -8 =-4a \Rightarrow a=2$$