Question

What is the horizontal asymptote of `f(x) = (x+1) / (x^2 +3x - 4)`?

Answer 1

y=0

Explanation:

If the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote. The degree is the power of the x variable(s).

Answer 2

`"horizontal asymptote at "y=0`

Explanation:

`"Horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" ( a constant)"`

`"divide terms on numerator/denominator by the highest"`
`"power of "x" that is "x^2`

`f(x)=(x/x^2+1/x^2)/(x^2/x^2+(3x)/x^2-4/x^2)=(1/x+1/x^2)/(1+3/x-4/x^2)`

`"as "xto+-oo,f(x)to(0+0)/(1+0-0)`

`rArry=0" is the asymptote"`
graph{(x+1)/(x^2+3x-4) [-10, 10, -5, 5]}