# Find max vertical distance

What is the maximum vertical distance between the line $y = x + 20$ and the parabola $y = x^2$ for $−4 ≤ x ≤ 5?$

What steps do I take to solve this? Do I have to use the distance formula and what do I do with the points it gave me?

If anyone could just bounce me in the right direction that would be neat. I can probably work an answer from there!

Also what's the distance formula to use here?

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The vertical distance at $x=a$ is the difference in $y$-coordinates at $x=a$, so it’s $|(x+20)-x^2|$. Now $x^2-x-20=(x+4)(x-5)$, so it’s negative between $x=-4$ and $x=5$. Thus, on the interval $[-4,5]$ we have $|(x+20)-x^2|=x+20-x^2$, not $x^2-x-20$.

Now let $f(x)=x+20-x^2$ and find the maximum of $f(x)$ on the interval $[-4,5]$.

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Draw a picture. Even though it is not necessary, note that the parabola and the line actually meet at $x=-4$ and $x=5$. Eyeball around where the maximum vertical distance might be.
The vertical distance, in our interval, is $(x+20)-x^2$. Maximize this in our interval, using whatever tools you prefer.
Maybe calculus. Or maybe note that $y=20+x-x^2$ is a downward facing parabola with vertex at $x=\frac{1}{2}$, so that value of $x$ gives the maximum distance. Or maybe complete the square.