MHB S5.t.5 Find domain and asymptotes.

[karush]

$\tiny{s5.t.5 Kiaser HS}$
Find domain asymptotes.
$g(x)=\dfrac{2x^2-14x+24}{x^2+6x-40}$
$\begin{array}{rll}
\textsf{factor}&=\dfrac{2(x-3)(x-4)}{(x-4)(x+10)}
=\dfrac{2(x-3)\cancel{(x-4)}}{\cancel{(x-4)}(x+10)}
=\dfrac{2(x-3)}{x+10}\\
\textsf{Domain} & -\infty<-10<\infty\\
HA \quad y&=3 \\
VA \quad x&=-10
\end{array}$
I think there is an oblique asymptote but ??
Also the OP has a hole at $x=4$ but they didn't ask for it ?

btw how come tab does not work here?

 

[skeeter]

what does $-\infty < -10 < \infty$ mean?
the domain is all reals except x = 4 and x = -10 …
$x \in (-\infty, -10) \cup (-10,4) \cup (4, \infty)$

horizontal asymptote is y = 2

oblique asymptotes occur when the degree of the numerator is one greater than the degree of the denominator

yes, the original rational function has a removable discontinuity at x = 4

 

[karush]

how do you get y=2 for HA

 

[skeeter]

how do you get y=2 for HA

$\displaystyle \lim_{x \to \pm \infty} \dfrac{2x^2-14x+24}{x^2+6x-40}$