Identify the vertical asymptote(s) of each function. Check all of the boxes that apply. f(x)=x-8/x^2-3x+2x = -8x = -2x = -1x = 1x = 2x = 8Nevermind the answer was x=1 and x=2. If anyone is wondering how to do it you can put the denominator in desmos and where the 2 points hit the x axis is your answer.

Since the function is:
[tex] \frac{x-8}{ x^{2} -3x+2} [/tex]

Now factorize the denominator:
[tex]\frac{x-8}{ x^{2} -x-2x+2}[/tex]

[tex]\frac{x-8}{ x(x-1) -2(x-1 ) }[/tex]

[tex]\frac{x-8}{ (x-1) (x-2 ) }[/tex]

Now put the denominator equals to zero:

[tex] (x-1) (x-2 ) = 0[/tex]

Now take each one of them separately to find the vertical asymptotes:
[tex]x-1 =0, x-2=0[/tex]

Ans: So there are TWO asymptotes:
1) x = 1.
2) x = 2.

-i