## lynn555 Group Title 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 one year ago one year ago

1. Jemurray3 Group Title

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

2. lynn555 Group Title

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

3. Jemurray3 Group Title

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

4. Jemurray3 Group Title

there should be no x involved.

5. LeeYeongKyu Group Title

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6. LeeYeongKyu Group Title

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. LeeYeongKyu Group Title

you can reduce it easily

8. Jemurray3 Group Title

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

9. lynn555 Group Title

ok thanks