Question

I'm stuck..

Answer 1

Question?

Answer 2

with?

Answer 3

me too

Answer 4

So, I always recommend drawing a picture to start, It gives you a ballpark

Answer 5

`So the distance to (2,1.5) is:`

Did you write that line? ^

Careful, it's (2, 0.5) right?

Answer 6

a point on the graph of \(y=x^2\) looks like
\[(x,x^2)\]
the square of the distance between that point ad \((2,\frac{1}{2})\) is
\[D^2=(x-2)^2+(x^2-\frac{1}{2})^2\]
you want to minimize that one

Answer 7

@zepdrix good point
where did the \(1.5\) come from?

Answer 8

k. I changed it to 0.5.

Answer 9

Also, think about how you minimize a function, what does that mean?

Answer 10

you mean find global minimum?

Answer 11

yes

Answer 12

good luck solving a cubic equation
hope it was cooked up to be not too bad

Answer 13

it's 5 /4 when x=1

Answer 14

yup to global min

Answer 15

oh yeah, was this ever cooked up
wait until you "simplify" the derivative
amazing!

Answer 16

yes it is

Answer 17

what am I supposed to do now?

Answer 18

be happy you are done
have a tequila

Answer 19

LOL

Answer 20

don't forget what \(x\) was, it was the first coordinate of the point on the graph

Answer 21

so point (1, 5/4)

Answer 22

Also, don't forget, when you find the x cooredinate of the global min, you need to put that into the original x^2 function

Answer 23

\(\frac{5}{4}\) is the minimum distance
actually it is the minimum of the square of the distance
the first coordinate of the point is \(1\) and the equation is \(y=x^2\) so you can find the second coordinate without difficulty

Answer 24

so 1,1? what was the point of finding the minimum then?

Answer 25

to find the \(x\) coordinate of the point!

Answer 26

how else you going to do it?
actually there is another way...

Answer 27

Quite true

Answer 28

minimizing the distance formula by finding the derivative makes your life wonderfully easy, you just have to remember to input the x value back into the original not use the value of the derivative :)