How to solve radical equation $y = \sqrt{9 x +18} - 3$?

Question

The question is

$$y = \sqrt{9 x +18} - 3$$ to solve this I would take 9 common I think and than go on with it. That is what I thought until I saw description about the transformation of my equation. It said vertical stretch of 3, left 2 and down one which makes no sense since if I had solved it by taking common I would have gotten 9(x +2) -3 so no stretch of 3 but nine. What did I do wrong?

Answer 1

HINT: Notice, given that $$Y=\sqrt{9x+18}-3$$ $$Y+3=\sqrt{9x+18}$$ Now, squaring both the sides we get $$(Y+3)^2=9x+18$$

Answer 2

$$Y = \sqrt{9 x +18} - 3$$

$$Y = \sqrt{9(x+2)} - 3$$

$$Y = \pm3\sqrt{x + 2} - 3$$

Not sure where the "down one" came from, but if $f(X) = \sqrt{X}$ then let $X = x + 2$ and you have

$$Y = \pm3f(x+2)-3$$

So $Y$ is $f$ stretched by 3, horizontaly shifted by -2, and vertically shifted -3.