Originally Posted by

**dwsmith** There is an error some where with how the derivative is affecting the linear system; however, I can't find it so I solved your equation a different way.

We know the slope of the tangent line is 0 at (1,7); therefore, the general equation is $\displaystyle \pm a(x-1)^2+7$.

If the parabola opens up, (-1,5) wouldn't be a point on the parabola.

So our equation becomes: $\displaystyle -a(x-1)^2+7$

$\displaystyle -a(-1-1)^2+7=5\rightarrow -4a+7=5\rightarrow -4a=-2\rightarrow a=\frac{1}{2}$

Now we have $\displaystyle \frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$

$\displaystyle \frac{-1}{2}+x-4$

$\displaystyle a=\frac{-1}{2}$, $\displaystyle b=1$ and $\displaystyle c=-4$