Perpendicular Asymptotes

From the denominator of $f$ we see immediately that $x = 2$ is the vertical asymptote of $f$ .

$g$ has the same vertical asymptote ( $x = 2$ ), so its denominator should be $x - 2$ .

We perform a division and rewrite $f$ as follows: $f(x) = \tfrac12x - 2 - \frac{1}{x-2}.$ This shows that $y = \tfrac12x - 2$ is the diagonal asymptote of $f$ . Its slope is $\tfrac12$ and it intersects that $y$ -axis in $(0, -2)$ .

The diagonal asymptotes are perpendicular . This translates into ( $\text{slope}_1\cdot\text{slope}_2 = -1$ ), so that the diagonal asymptote of $g$ has slope $-2$ .

The asymptotes intersect on the $y$ -axis . The diagonal asymptote of $g$ must therefore also go through $(0, -2)$ . Combined with the previous observation, we now know that the diagonal asymptote of $g$ is the line $y = -2x -2.$

The function $g$ has the same pattern as $f$ , and can also be written in the form $g(x) = -2x -2 + \frac{\lambda}{x-2}.$

The graphs intersect at $x = -2$ . Now $f(-2) = -2\tfrac34$ , so that $g(-2) = -2\cdot (-2)-2+\frac{\lambda}{(-2)-2} = 2 - \frac{\lambda}{4} = -2\tfrac34.$ The solution is $\lambda = 19$ . We now know the function $g$ : $g(x) = -2x - 2 + \frac{19}{x-2} = \frac{-2x^2+2x+23}{x-2}.$ (There is really no need to calculate the last expression--it's just in case you're curious about $a, b, c, d$ .)

Finally, we need to find the other intersection point. Equate the two functions: $f(x) = g(x) \\ \tfrac12x - 2 - \frac{1}{x-2} = -2x - 2 + \frac{19}{x-2} \\ 2\tfrac12x = \frac{20}{x-2} \\ 2\tfrac12x^2 - 5x - 20 = 0 \\ x^2 - 2x - 8 = 0 \\ (x+2)\cdot (x-4) = 0 \\ x = -2\ \ \text{or}\ \ x = \boxed{4}.$

• The vertical asymptote 'happens where' the function value 'blows up' as the denominator gets very small. For f this is at $x=2$ and for g this is at $x=d$ . The fact that the vertical asymptotes of f and g coincide tells us that $d=2$ .

• The diagonal (slant/oblique) asymptote is found by doing a long division.

$f(x) = \frac{ x^2-6x+6}{2x-4} = \frac{x}{2} - 2  -\frac{1}{x-2}$ so that the slant asymptote of f is given by the line $y= \frac{x}{2}-2$ . Similarly $g(x)=\frac{ax^2+bx+c}{x-d}=ax+b+ad+\frac{c+bd-ad^2}{x-d}$ so that the slant asymptote of g is given by the line $y=ax+b+ad$ .

The slope of the asymptote of f equals $\frac{1}{2}$ and that of the asymptote of g equals $a$ . The fact that these are perpendicular tells us that $a=-2$ .

The fact that both asymptotes intersect at the y-axis tells us (by putting x=0) that $-2=b+ad$ , so

$b=4-2=2$ .

• To study the intersection of the graphs of f and g we set f=g and cross multiply to remove the fractions:

$(x^2-6x+6)(x-d)=(ax^2+bx+c)(2x-4)$ . Set $x=-2$

$(4+12+6)(-2-2)=(4a-2b+c)(-4-4)$ . Fill in the knowns $a=-2, b=2, d=2$ and divide by $-8$

$11=-8-4+c$

$c=23$

Now that g is known we repeat this exercise with known g but unknown x, in order to find the other intersection point. $\frac{x^2-6x+6}{2x-4}=\frac{-2x^2+2x+23}{x-2}$ . Multiply by a factor $2x-4$ .

$x^2-6x+6 = -4x^2+4x+46$

$5x^2-10x-40=0$

$x=\frac{10 \pm \sqrt{100+800}}{10}$

$x=1 \pm 3$

$x=-2 \vee x=\boxed{4}$