Question

State the vertical asymptote of the rational function.
f(x)= (x-3)(X+4)/x^2-1

Answer 1

Factor the denominator.
\[x^2 - 1 = (x + 1)(x - 1)\]Now set each factor equal to 0 and solve for x.
The reason this works is because there are vertical asymptotes when the denominator is equal to 0.
x + 1 = 0
x - 1 = 0
x = ?, ?

Answer 2

but how does \[x ^{2}-1=(x+1) (x-1)\] ?

Answer 3

It's the difference in squares pattern.
\[(a^2 - b^2) = (a + b)(a - b)\]

Answer 4

alright thank you

Answer 5

np :)