Function with Properties

Question

Let $$f(x) = \frac{x^2 - 6x + 6}{2x - 4} $$and $$g(x) = \frac{ax^2 + bx + c}{x - d}.$$You are given the following properties:

$\bullet$ The graphs of $f(x)$ and $g(x)$ have the same vertical asymptote.

$\bullet$ The oblique asymptotes of $f(x)$ and $g(x)$ are perpendicular, and they intersect on the $y$-axis.

$\bullet$ The graphs of $f(x)$ and $g(x)$ have two intersection points, one of which is on the line $x = -2.$

Find the point of intersection of the graphs of $f(x)$ and $g(x)$ that does not lie on the line $x = -2.$

So far, I've graphed the function with most of the conditions given, but not sure how to go after that.

Answer 1

• vertical asymptote means denominators have the same root

So $d=2$

• oblique asymptote means higher factor when $x\to\infty$

$f(x)\sim \frac x2$ so we need $g(x)\sim -2x\ $ for the lines to be perpendicular.

So $a=-2$

• oblique asymptotes cross on $y$-axis

We can do partial decomposition now, to get the equations of asymptotes

$f(x)=\underbrace{(\frac 12 x-2)}_\text{oblique asymptote}-\frac 1{x-2}$

$g(x)=\underbrace{(-2x+b-4)}_\text{oblique asymptote}+\frac {c+2b-8}{x-2}$

$f$ asymptote cross $y=0$ at $x=4$ and value of $g$ asymptote is $y=b-12$ there.

So $b=12$

• common point in $x=-2$

$f(-2)= -\frac {11}4 = g(-2) = 8 -\frac c4$

So $c=43$

• another common point

Solve $f(x)=g(x)\implies x=8$