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I have got a very simple problem. I have an exercise:

If $\ f(x) = 2x^2 − 4$, give the function which shows the graph of $\ f(x)$ after vertical stretch of scale factor $\ 0.5 $ followed by a translation $\binom{-4}{0}$

The answer that I get is $\ f(x)=x^2+8x+14 $, but the answer given is $\ f(x)=8x^2+64x+124 $. In my opinion, the answer that is given can certainly be achieved, but using horizontal translation instead of vertical. After drawing a graph of my function and the given function I noticed that in my case the function is compressed (its "branches" are closer to the x - axis than the original one) - as it should be, as scale factor is less than 1.

Am I wrong there, or is something wrong with answers? I would not have asked the question, but I noticed that there is at least one more question about which I am uncertain as much as about this one, thus, I need to find out the real answer.

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    $\begingroup$ $x^2+8x+14=((x+4)^2-4)/2$ is correct. Reusing the symbol $f(x)$ to mean different things leads to confusion. $\endgroup$
    – user574889
    Jul 17, 2018 at 13:49
  • $\begingroup$ @cactus, sorry, did not think about changing it to something else. Thanks for the answer! That means that there are 3 mistakes in a row in the exercises... $\endgroup$
    – Danielius
    Jul 17, 2018 at 13:52

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You are right.

We are looking for the function \begin{align}\frac12f(x+4)&=\frac12(2(x+4)^2-4)\\&=(x+4)^2-2\\ &=x^2+8x+14 \end{align}

Of course, there is a possibility that you course is asking the wrong question as well.

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  • $\begingroup$ Thanks for the answer! That means that there are 3 mistakes in a row in those exercises... I will accept the answer soon. $\endgroup$
    – Danielius
    Jul 17, 2018 at 13:53
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$$f(x)=2(x^2-4)$$ Vertical stretching with scale $1:2$: $$f_1(x)=0.5 f(x)=x^2-4$$ Translation by a vector $[-4,0]$: $$f_2(x)=f_1(x-(-4))+0=(x+4)^2-2=x^2+8x+14$$

Thus your solution seems to be fine.

The textbook solution might be created by taking a stretching factor $4$ and translation $[-4,12]$

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