Answer
See image
Work Step by Step
The graphs of $f$ and $f^{-1}$ (if the iverse is defined, )
are symmetrical about the line $y=x.$
Symmetry:
Reflection of the point (a,b) about the line $y=x$ is the point (b,a).
The graph of $f(x)=\displaystyle \frac{1}{4}x+2$ is an oblique line, passes the horizontal line test,
so $f^{-1}(x)$ exists.
Graph f:
We find two points needed to graph a line:
$ x=0 \Rightarrow f(0)=2\qquad$...$ (0,2)$ is on the graph of f.
$ x=4 \Rightarrow f(4)=3\qquad$...$ (4,3)$ is on the graph of f.
Plot the points and join with a straight line - we have the graph of f.
Graph $f^{-1}:$
Use the symmetry:
$(2,0)$ is on the graph of $f^{-1}(x)$ and
$(3,4)$ is on the graph of $f^{-1}(x)$
Plot the points and join with a straight line - we have the graph of $f^{-1}$.
