# Thread: Point on the graph closet to set coords.

1. ## Point on the graph closet to set coords.

Which point on the graph

$y=4x^3+2$

closest to 2,0

I just need a general idea of what to do and Ill do the work

2. Originally Posted by sk8erboyla2004
Which point on the graph

$y=4x^3+2$

closest to 2,0

I just need a general idea of what to do and Ill do the work
The general idea is illustrated in these threads:

http://www.mathhelpforum.com/math-he...e-problem.html

http://www.mathhelpforum.com/math-he...-x-origin.html

3. Ok If I use the distance forumla I get

$D=\sqrt{(x-2)^2+(4x^3+2)^2}$

If I simplify the inside which I assume I can I get

$D=\sqrt{16x^6+17x^2+8}$

?

Which when I do the derivative of

$\frac{96x^5+34x}{2\sqrt{16x^6+17x^2+8}}$

then I solve the numerator for 0? which I dont know how I would do that

4. Originally Posted by sk8erboyla2004
Ok If I use the distance forumla I get

$D=\sqrt{(x-2)^2+(4x^3+2)^2}$

If I simplify the inside which I assume I can I get

$D=\sqrt{16x^6+17x^2+8}$

?

Which when I do the derivative of

$\frac{96x^5+34x}{2\sqrt{16x^6+17x^2+8}}$

then I solve the numerator for 0? which I dont know how I would do that
$96x^5+34x=0 \iff x(96x^4+34)=0$

So either $x=0 \mbox{ or } 96x^4+34=0$

so x=0. we reject the other because it only has complex roots.

5. So all steps taking the derivative where correct?

If so much appreciated either way!

So if that is correct the point closet to graph is

(2,0) ?

6. Originally Posted by sk8erboyla2004
So all steps taking the derivative where correct?

If so much appreciated either way!

So if that is correct the point closet to graph is

(2,0) ?
Yes.

The numerator was correct. Just as a note in this type of problem you can just minimize the square of the distance .

7. Ok If I have this

$2x-3y=12$ closet to (7,8)

I solve for y

I get $y=\frac{12-2x}{3}$

so would my distance formula look like this?

$\sqrt({x-7})^2+({\frac{12-2x}{3}-8})^2$

?

8. Originally Posted by sk8erboyla2004
Ok If I have this

$2x-3y=12$ closet to (7,8)

I solve for y

I get $y=\frac{12-2x}{3}$

so would my distance formula look like this?

$\sqrt({x-7})^2+({\frac{12-2x}{3}-8})^2$

?
Almost- you have the square root only around the $(x-7)^2$. It should be around the entire expression. You do that by putting the entire expression in { }. But as TheEmptySet said earlier, to minimize $\sqrt{f(x)}$, it is sufficient to minimize f(x). So find the x and y that minimize $(x- 7)^2+ \left(\frac{12- 2x}{3}- 8\right)^2$.

9. So differentiate for x and and find y