Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.