Here's the question you clicked on:
lynn555
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
\[L = \sqrt{ (x-4)^2 + y^2 } \] and \[ y = \sqrt{x+3} \implies x = y^2 - 3\] so substitute that in.
so the answer would be y^5-46+x?
erm, no. Literally substitute y^2-3 in for x in the equation that starts with L = ...
there should be no x involved.
|dw:1347425169078:dw|
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
you can reduce it easily
That should be y^2-7, not y^2-6, and the whole thing should be under a square root. But yeah.