Maximum vertical distance between the line $y = x + 30$ and the parabola $y = x^2$ for $−5 ≤ x ≤ 6$

Question

What is the maximum vertical distance between the line $y = x + 30$ and the parabola $y = x^2$ for $−5 ≤ x ≤ 6$?

This is what I did but didn't work:

Set $y_1=x+30$ and $y_2=x^2$, plugged those into the distance formula [assuming $x=a$, $d=\sqrt{(y2-y1)^2}$ and get $d(x)=x^2-x-30$.

Assuming this is correct, I found the derivative of the equation to determine $x$ and then used the $x$ into $d(x)=x^2-x-30$ to find $y$ but my answer is wrong. Where is my mistake?

Answer 1

Hint: Maximize the (quadratic!) function $f(x) = x+30-x^2$ on $[-5,6]$.