consider the point (x,y) lying on the graph of y=(x+3)^1/2. let L be the distance between the points (x,y) and (4,0). write l as a function of y

Question

anonymous · Answer

$$L  = \sqrt{ (x-4)^2 + y^2 } $$
and
$$ y = \sqrt{x+3} \implies x = y^2 - 3$$
so substitute that in.

anonymous · Answer

so the answer would be y^5-46+x?

anonymous · Answer

erm, no.  Literally substitute y^2-3 in for x in the equation that starts with L = ...

anonymous · Answer

there should be no x involved.

anonymous · Answer

|dw:1347425169078:dw|

anonymous · Answer

so given function is y=(x+3)^1/2 so y^2=x+3, so, x = y^2-3.

plugging x to the L=(x-4)^2 + y^2

then, L=(y^2-6)^2 + y^2

anonymous · Answer

you can reduce it easily

anonymous · Answer

That should be y^2-7, not y^2-6, and the whole thing should be under a square root.  But yeah.

anonymous · Answer

ok thanks