Mathematics OpenStudy (anonymous):

The point B(a,b) is on the curve f(x)=x² such that B is the point which is closest to A(6,0). Calculate the value of a. OpenStudy (anonymous):

a point on the curve $$y=x^2$$ looks like $$(x,x^2)$$ and distance between $$(x,x^2)$$ and $$(0,6)$$ is by the distance formula $\sqrt{(x-0)^2+(x^2-6)^2}$ easier to work with the square of the distance, so minimize $x^2+(x^2-6)^2$ OpenStudy (anonymous):

i am guessing this is calc, so multiply out to get $d=x^4-11 x^2+36$ then maybe take the derivative to find the critical points good from there? OpenStudy (anonymous):

Oh, shoot, I'm sorry the question said A(6,0) not (0,6). OpenStudy (anonymous):

should be easy now since $d'=4x^3-22x=2x(x^2-11)$ zeros are easy and they will give you a minimum OpenStudy (anonymous):

But it's the same, right? Just swap the numbers in the distance formula. OpenStudy (anonymous):

in that case it is even easier OpenStudy (anonymous):

yeah it is the same, just swap OpenStudy (anonymous):

$d^2=(x-6)^2+x^4=x^4+x^2-12 x+36$ OpenStudy (anonymous):

oh actually now it is not easier. you have to find the zeros of a cubic function. i liked the last one more OpenStudy (anonymous):

we can do it a different way, but you get exactly the same equation to solve, so it will be no easier OpenStudy (anonymous):

Well, I am allowed to use a calculator. So I think the cubic equation won't be a problem. :) OpenStudy (anonymous):

well then we can go right to the answer http://www.wolframalpha.com/input/?i=%28x-6%29^2%2Bx^4 OpenStudy (anonymous):

So since local min at x=1.33, is that the answer? OpenStudy (anonymous):

yeah looks like that OpenStudy (anonymous):

Ok thanks! :) OpenStudy (anonymous):

yw

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