Answer:
Asymptotic discontinuities at [tex]x = (-2)[/tex] and [tex]x = 4[/tex].
Step-by-step explanation:
A linear function has an asymptotic discontinuity at [tex]x = a[/tex] if [tex](x - a)[/tex] is a factor of the denominator after simplification.
The numerator of this function, [tex](x - 1)[/tex], is linear in [tex]x[/tex].
The denominator of this function, [tex](x^{2} - 2\, x - 8)[/tex], is quadratic in [tex]x[/tex]. Using the quadratic formula or otherwise, factor the denominator into binominals:
[tex]\begin{aligned}y &= \frac{(x - 1)}{x^{2} - 2\, x - 8} \\ &= \frac{(x - 1)}{(x - 4)\, (x + 2)}\end{aligned}[/tex].
Simplify the function by liminating binomials that are in both the numerator and the denominator.
Notice that in the simplified expression, binomial factors of the denominator are [tex](x - 4)[/tex] and [tex](x + 2)[/tex] (which is equivalent to [tex](x - (-2))[/tex].) Therefore, the points of discontinuity of this function would be [tex]x = 4[/tex] and [tex]x = (-2)[/tex].
