# Question #22332

Feb 13, 2015

The answer is E) $\frac{5}{4}$.
We start with the fact that graphs of any function and its inverse are symmetrical relatively to the angle bisector of the main angle of the first quadrant, that is relatively to a straight line described by an equation $y = x$. This and many other properties of graphs are discussed in details in the course of Advanced Math for Teens at the Web site Unizor in the chapter Algebra - Graphs.

Therefore, the points of intersection between the graph of $y = f \left(x\right) = {\left(x - 1\right)}^{2} + k$ and its inverse $y = {f}^{- 1} \left(x\right)$ lie on this angle bisector $y = x$.

To determine these points of intersection, we have to solve the equation
$x = {\left(x - 1\right)}^{2} + k$
This is a simple quadratic equation
${x}^{2} - 3 x + k + 1 = 0$
It's solutions are
${x}_{12} = \left(\frac{1}{2}\right) \cdot \left(3 \pm \sqrt{9 - 4 \left(k + 1\right)}\right)$

As $k$ increases, there will be a point when the expression under the square root becomes and stays negative. Up to that moment we can increase $k$. Therefore, the maximum $k$ is the one when
$9 - 4 \left(k + 1\right) = 0$,
that is $k + 1 = \frac{9}{4}$ and $k = \frac{5}{4}$