# Steps of transformation

Given the function $y=-5-3 \sqrt{-2x-4}$ and base function y= $\sqrt{x}$ describe the transformations that have been applied to obtain the function from the base function.

I tried, horizontal compression by factor of half, vertical stretch by factor of 5, reflection in both x and y axis, translation 2 units to to the left and finally, translation 3 units down. I'm not sure if I'm doing it right, HELP!!

• I tried, horizontal compression by factor of half, vertical stretch by factor of 5, reflection in both x and y axis, translation 2 units to to the left and finally, translation 3 units down. I'm not sure if I'm right, though. Could you please help me out? Thanks in advance Nov 8, 2015 at 4:09
• Alright. Could you help me with the question please? Nov 8, 2015 at 4:17

I like to work my way from inside to outside. We're given $$y=-5-3 \sqrt{-2x-4} = -5-3 \sqrt{-2(x+2)}$$

Thus the inside most transformation is $\sqrt{x} \mapsto \sqrt{x+2}$. This shifts the function to the left by $2$.

The next inside most transformation is $\sqrt{x+2} \mapsto \sqrt{-2(x+2)}$. This corresponds to reflecting the graph through the $y$-axis and compressing it in the horizontal direction by a factor of $2$.

Then comes $\sqrt{-2(x+2)} \mapsto -3\sqrt{-2(x+2)}$. This reflects the graph through the $x$-axis and stretches it in the vertical direction by a factor of $3$.

And finally $-3\sqrt{-2(x+2)} \mapsto -5-3\sqrt{-2(x+2)}$. This shifts the graph downward by $5$.

So $\sqrt{x} \mapsto -5-3\sqrt{-2(x+2)}$ shifts the function left by $2$, then reflects across the $y$-axis and compresses in the horizontal direction by a factor of $2$, then reflects across the $x$-axis and stretches by a factor of $3$, and finally shifts down by $5$.

• Thank you so much! It was really nice of you to explain everything so throughly, you really cleared my doubt. Thank you, once again. Nov 8, 2015 at 4:29
• No problem. :-) Don't forget to accept my answer by clicking on the checkmark to the left of it (it'll turn green once you do).
– user137731
Nov 8, 2015 at 4:33
• It's my first time using this website so, I really had no idea I was suppose to do that. You're really smart! Nov 8, 2015 at 4:39