Given that $g(x) = \sqrt{1-x}$, I'm asked to graph $-g(-3x+1)$. In order to do so, I first listed out the necessary transformations as follows:
(1) Let $a(x) = -g(x)$. Then, if $(x,y)$ is in the graph of $g$, it follows that $a(x) = -g(x) = -y$ implies that $(x,-y)$ is on the graph of $a$ which implies that the graph of $-g$ is reflected over the $x$-axis.
(2) Let $b(x) = a(3x)$. Then, $b\bigg(\dfrac{x}{3}\bigg)=a(x)=-g(x)$. Thus, if $(x,y)$ is on the graph of $a$, then $\bigg(\dfrac{x}{3},y\bigg)$ is on the graph of $b$. So, the graph of $a$ is contracted horizontally by a factor of $3$ towards the $y$-axis.
(3) Let $c(x) = b\bigg(x+ \dfrac{1}{3} \bigg) = a\bigg(3\bigg(x + \dfrac{1}{3}\bigg)\bigg) = a(-3x+1) = -g(-3x+1)$. Then, $c\bigg(x - \dfrac{1}{3} \bigg) = b(x) = a(3x) = -g(3x)$. This means that each point on the graph of $c$ is shifted to the left by $\dfrac{1}{3}$ unit.
(4) Let $d(x) = c(-x) = b\bigg(-x + \dfrac{1}{3} \bigg) = a\bigg(\bigg(-x+ \dfrac{1}{3}\bigg)\bigg) = a(-3x+1) = -g(-3x+1)$. Then, if $(x,y)$ is on the graph of $c$, it follows that $d(-x)=c(x)$ implies $(-x,y)$ is on the graph of $d$. So, $c$ is reflected over the $y$-axis.
I was wondering if there's a simpler way to go about graphing $-g(-3x+1)$? If so, should I be going from step (4) to step (1)?
In general, the transformations to get from $y=f(x)$ to $y=af(k(x-c))+d$ are listed as follows:
As such, to get from $g(x)$ to $-g(-3x+1)=-g(-3(x-1/3))$, we have: