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If the base function for the transformed function $$f(x)= -4(3x)^2+5$$ is $f(x)=x^2$, then: is $k=3$ or is $k=9$?

By comparing the transformed function to: $af(k(x-d))+c$, you can pinpoint the factors which affect the function. In this case, $a=-4, k=3$ or $k=9$, $d=$(doesn't apply) and $c=5$. My notes imply that for the term $(3x)^2$, $k=3$ which represents a horizontal compression by $1/3$ units. But this would apply to $3x^2$, which is a totally different expression. I therefore think that $k=9$, which represents a horizontal compression of the function by $1/9$ units. What do you all think? Thanks in advance.

Haris
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Stephen Makanga
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – cpiegore Feb 27 '24 at 03:01
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    Give us more context - what does the parameter $k$ mean in general? It may mean something else for others – Sean Roberson Feb 27 '24 at 03:03
  • $$k=3$$

    $$-4(3(x-0))^2+5$$

     – Lucien Jaccon Feb 27 '24 at 04:27
  • Thanks. However, my base function is y=x^2, right? So all transformations are applied to x^2. Therefore, I still think that applying k to only 3x in (3x)^2 is applying the factor to a much different expression. – Stephen Makanga Feb 28 '24 at 16:08

1 Answers1

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Notice that in the given expression for $f(x)$, all of $3x$ is being squared. Which means that $3x$ is the term that is being input to $g(x) = x^2$, so our transformation on $x$ is $x \rightarrow 3x$.

This does represent a horizontal compression, because the value we would have associated with $x = 1$ we now associate with $3x = 1 \implies x = \frac{1}{3}$.

By comparison, the $-4$ being multiplied out the front gives us a flip and a vertical expansion, because whatever comes out of $g(x)$ we are now scaling up by a factor of 4 and reflecting in the $y$-axis.

With that said, if you're not tied to the exact formatting of the expression, there are multiple valid choices for $a$ and $k$ in this particular case, as long as $ak^2 = -36$.

ConMan
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