Answer
$ a.\quad$
$f^{-1}(x)=-x+1$
Intersects the line $y=x$ at a right angle.
See image below.
$ b.\quad$
$f^{-1}(x)=-x+b$
Intersects the line $y=x$ at a right angle.
$ c.\quad$
The function is its own inverse.
Work Step by Step
$ a.\quad$
Find the inverse:
Set $y=f(x)$. Interchange $x$ and $y$:
$y=-x+1$
$x=-y+1$
... solve for y
$x-1=-y$
$y=-x+1$
... replace y with $f^{-1}(x)$
$f^{-1}(x)=-x+1$
See image for graphs.
The slope of $f(x) $ is $m_{1}=-1$; the slope of the line $y=x $ is $m_{2}=1$
Perpendicular lines have slopes for which $m_{1}\cdot m_{2}=-1.$
(The angle of intersection is a right angle.)
This can also be deduced as in the image, using a square of side 1.
$ b.\quad$
The inverse of $f(x)=-x+b$ is found by
$ y=-x+b\qquad$ ... replace f(x) with y
$ x=-y+b\quad$ ... interchange x, ad y. Solve for y
$x-b=-y$
$y=-x+b$
$f^{-1}(x)=-x+b$
Again, this line has slope $-1$, and is perpendicular to the line $y=x,$ which has slope 1.
$ c.\quad$
If the graph of the function is perpendicular to the line $y=x,$ the graph of the inverse is the same as the graph of the original function. The function is its own inverse.
