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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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3 Answers 3

up vote 2 down vote accepted

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.

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+1: Nice alternatives to using calculus. –  Cameron Buie Jun 30 '12 at 18:17

Instead of maximizing the vertical distance, it is convenient to maximize the squared vertical distance $d^2(x)=(x^2-x-20)^2$.

We cancel the first derivative to find the extrema, $\left(d^2(x)\right)'=2d(x)d'(x)=0$.

In this product, when $d(x)$ cancels the distance is $0$ and corresponds to the global minimum, which we can ignore.

Now $d'(x)=2x-1=0$ is the only maximum, such that $d(\frac12)=\frac{81}4$.

But for completeness, we must also evaluate the distance at the domain endpoints,

$$d(-4)=0\text{, and }d(5)=0,$$ showing that the seeked maximum is indeed $\frac{81}4$.

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