Question

describe the vertical asymptote(s) and hole(s) for the graph of y=x-4/x^2+3x+2

Answer 1

the degree of the numerator is less than the degree of the denominator, so you have a horizontal asymptote at \(y=0\)

Answer 2

i am assuming this is
\[y=\frac{x-4}{x^2+3x+2}\]right?

Answer 3

Yes

Answer 4

to find the vertical asymptotes, set the denominator equal to zero and solve for \(x\(
i.e. solve
\[x^2+3x+2=0\]

Answer 5

\[x^2+3x+2=0\]

Answer 6

this one factors so it should be easy enough
\[(x+2)(x+1)=0\] etc

Answer 7

then you will set each of them equal to zero right?