Horizontal transformations

Any graph feature (asymptote, intercept etc) on the graph of y=f\left(x\right) will be horizontally transformed due to the 3-2x argument of the new graph y=-3f\left(3-2x\right). If the original graph has a certain feature at x=k, then the new graph will have that same feature when

3-2x=k\Rightarrow x=\frac{3-k}{2}

So the vertical asymptotes x=0 and x=-3 of y=f\left(x\right) become

x=\frac{3-0}{2}=\frac32
x=\frac{3-\left(-3\right)}{2}=3

The x-intercepts of f\left(x\right)=0 (x=-\frac32 and x=\frac12) are horizontally transformed, but not vertically since -3f\left(3-2x\right)=-3\cdot0=0 (there is no vertical shift, and scaling 0 gives 0). Therefore the new graph will have x-intercepts at

x=\frac{3-\left(-\frac32\right)}{2}=\frac94
x=\frac{3-\frac12}{2}=\frac54

Vertical transformations

The horizontal asymptote at y=2 will become a horizontal asymptote at y=-3\cdot2=-6 .

Meanwhile, the labeled point \left(3,\frac53\right) on the graph of y=f\left(x\right) will be transformed horizontally:

3-2x=3\Rightarrow x=0

And vertically:

y=-3\cdot\frac53=-5

So there will be a y-intercept at -5 .

Problem Bank - Transformations & asymptotes

Problem Bank - Transformations & asymptotes