6, 8, 3, 10 will be the answer for this question
Answer: y=x^8-x^7+x^6-x^5+x^4+x^3+x^2+3
Step-by-step explanation:
here is an example function that satisfies the requirement:
y=x^8-x^7+x^6-x^5+x^4+x^3+x^2+3
if you plug in x=0,
you get y= 3, which satisfies (0,3)
if you plug in x=1,
you get y = 1 - 1 + 1 - 1 + 1 + 1 + 1 + 3,
you get y=6, which satisfies (1,6)
The pair of limits that would verify the existence of a vertical asymptote at x = –1 is:
![\lim_{x \rightarrow -1^-} = \infty, \lim_{x \rightarrow -1^+} = \infty](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Crightarrow%20-1%5E-%7D%20%3D%20%5Cinfty%2C%20%5Clim_%7Bx%20%5Crightarrow%20-1%5E%2B%7D%20%3D%20%5Cinfty)
<h3>What are the vertical asymptotes of a function f(x)?</h3>
The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
For a vertical asymptote at x = -1, the denominator is zero at -1, hence both lateral limits at x = -1 go to infinity, that is:
![\lim_{x \rightarrow -1^-} = \infty, \lim_{x \rightarrow -1^+} = \infty](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Crightarrow%20-1%5E-%7D%20%3D%20%5Cinfty%2C%20%5Clim_%7Bx%20%5Crightarrow%20-1%5E%2B%7D%20%3D%20%5Cinfty)
More can be learned about vertical asymptotes at brainly.com/question/27949428
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