For this question, we'll need the vertex form of a parabola, which, for a quadratic function, is:
![y=a(x-h)^2+k](https://tex.z-dn.net/?f=y%3Da%28x-h%29%5E2%2Bk)
where (h,k) is the vertex of the parabola. In this case, our parabola is opening horizontally to the right, so we need to swap the x and y in our equation. We have:
![x=a(y-h)^2+k](https://tex.z-dn.net/?f=x%3Da%28y-h%29%5E2%2Bk)
We've been given a the coordinates of the vertex (-4,-1) as well as the coordinates for a point on the parabola (2,0), so we can substitute in the values from both of these coordinates to easily solve for a, the coefficient of the squared term.
![h=-4\\k=-1\\x=2\\y=0\\\\2=a(0-(-1))^2+(-4)\\2=a(1)^2-4\\2=a-4\\6=a](https://tex.z-dn.net/?f=h%3D-4%5C%5Ck%3D-1%5C%5Cx%3D2%5C%5Cy%3D0%5C%5C%5C%5C2%3Da%280-%28-1%29%29%5E2%2B%28-4%29%5C%5C2%3Da%281%29%5E2-4%5C%5C2%3Da-4%5C%5C6%3Da)
So, the coefficient of the squared term for this parabola is 6.