Find vertical asymptotes

f(x) = x + 1/x^2 - 8x + 7

I know it's 1,7 but I have no idea how they got that. can someone help?

Vertical asymptotes will occur when the denominator equals zero when the numerator doesn't.

By setting the denominator equal to zero, we have:
x^2 - 8x + 7 = 0
==> (x - 7)(x - 1) = 0, by factoring
==> x = 1 and x = 7, by the zero-product property.

Since the numerator equals 2 and 8 when x = 1 and x = 7, respectively, we see that the VAs are x = 1 and x = 7.

I hope this helps!

It 1 and 7 are the correct answers, you must have left some parentheses out of the problem statement.

I assume
x + 1/x^2 - 8x + 7
is actually
(x+1)/(x²-8x+7)
instead of
x + (1/x²) - 8x + 7

If that assumption is correct then the vertical asymptotes are the vertical lines through the values of x where the denominator equals zero:
(x²-8x+7) = 0
(x-7)(x-1) = 0
x = 1, 7

f(x) = [ (x + 1) ] / [(x^(2) - 8x + 7) ]

To find vertical asymptotes of the f(x) function look for the x - values in the denominator which make the function undefined.

Therefore in order to do that you need to factor the denominator.

f(x) = [(x + 1) ] / [ (x - 7) * (x - 1)]

So set x - 7 = 0, x - 1 = 0

x - 7 = 0

Add 7 to both sides.

x = 7

x - 1 = 0

Add 1 to both sides.

x = 1

Therefore the vertical asymptotes are at x = 1 and x = 7 -----------> ANSWER