## anonymous one year ago Help with inverse equation h(x)= 2 sqrt. (x+3) x= 2 sqrt. (y+3) x^2= 2 sqrt. (y+3)^2 x^2= 2y+6 -6. -6 x^2 -6= 2y ÷2. ÷2 (x^2 -6)/2= y ?????? The answer should be 1/4 x-3 I cannot get the 1/4 help please!

1. anonymous

You need to get rid of the radical sign $$\sqrt{y+3}$$, so you must square it ( $$(\sqrt{y+3})^2$$) to get $$y+3$$. But of course, you must square it on both sides! So in the left side, you'll get $$\frac{ x^2 }{ 4 }$$ or $$\frac{ 1 }{ 4 }x^2$$.

2. anonymous

Once you've done that, you can now easily solve for y.