Question

Need help!!!

Find a quadratic function whose x-intercepts are -4 and 2 and whose range is [-18, 00).

Answer 1

So we know that it has x-intercepts of -4 and 2 and that it's vertex is in between those two at a y-value of -18!

The average of the two is:
\[x_{av}=\frac{x_1+x_2}{2}=\frac{-4+2}{2}=\frac{-2}{2}=-1\]
So then, we know that the vertex is located at the point of \(P(-1,-18)\)
In general, for a parabola with the vertex located at the point \(V(h,k)\), the formula is:
\[f(x)=a(x-h)^2+k\]
So we know:
\(h=-1\)
\(k=-18\)
And we can simplify the equation:
\[\eqalign{
&f(x)=a(x-(-1))^2+(-18) \\
&f(x)=a(x+1)^2-18
}\]
Now we need to find \(a\). Following so far?

Answer 2

yes

Answer 3

but how do we find a?