Exponential & Logarithmic Parent Function
2. The parent function 𝑓(𝑥) = 𝑙𝑜𝑔3 𝑥 is vertically compressed by a factor of 1/4, reflected about the x-axis, horizontally translated horizontally c units, vertically translated 4 units up, and y-intercept is 3.
a) What is the value of c?
b) What is the domain of the transformed function?
1 Answer
Consider the parent function $𝑓(𝑥) = \log3x$.
- vertically compressed by a factor of 1/4 $\Rightarrow$ $𝑓_1(𝑥) = \frac{1}{4}\log3x$
- reflected about the x-axis $\Rightarrow$ $𝑓_2(𝑥) = -\frac{1}{4}\log3x$
- horizontally translated horizontally c units $\Rightarrow$ $𝑓_3(𝑥) = -\frac{1}{4}\log3 (x-c)$
- vertically translated 4 units up $\Rightarrow$ $𝑓_4(𝑥) = -\frac{1}{4}\log3 (x-c)+4$
(a) Since the y-intercept is 3 we have
\[3=𝑓_4(0) = -\frac{1}{4}\log3 (0-c)+4\]
so
\[-\frac{1}{4}\log (-3 c)=3-4=-1 \Rightarrow \log (-3c)=4\]
\[-\frac{1}{4}\log (-3 c)=3-4=-1 \Rightarrow \log (-3c)=4\]
Hence
\[-3c=e^4.\]
So $$c=\frac{e^4}{-3}.$$
(b) We have
$$𝑓_4(𝑥) = -\frac{1}{4}\log3 (x-c)+4=-\frac{1}{4}\log(3 x+e^4)+4.$$
So we should have $3x+e^4 >0$, and hence
\[x> -\frac{e^4}{3}.\]
So the domain of the function is $(-\frac{e^4}{3}, \infty)$.

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