## anonymous 3 years ago 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

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

there should be no x involved.

5. anonymous

|dw:1347425169078:dw|

6. anonymous

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

7. anonymous

you can reduce it easily

8. anonymous

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

9. anonymous

ok thanks