1
$\begingroup$

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!!

Thanks in advance

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – Salik
    Nov 8, 2015 at 4:09
  • $\begingroup$ Alright. Could you help me with the question please? $\endgroup$
    – Salik
    Nov 8, 2015 at 4:17

1 Answer 1

0
$\begingroup$

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 $5$.

$\endgroup$
3
  • $\begingroup$ Thank you so much! It was really nice of you to explain everything so throughly, you really cleared my doubt. Thank you, once again. $\endgroup$
    – Salik
    Nov 8, 2015 at 4:29
  • $\begingroup$ 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). $\endgroup$
    – user137731
    Nov 8, 2015 at 4:33
  • $\begingroup$ It's my first time using this website so, I really had no idea I was suppose to do that. You're really smart! $\endgroup$
    – Salik
    Nov 8, 2015 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.