# How do you find the domain, identify any horizontal, vertical, and slant (if possible) asymptotes and identify holes, x-intercepts, and y-intercepts for (x^2-1)/(x^2+4)?

Nov 16, 2017

Domain = $\mathbb{R} \in \mathbb{N}$
H.A. = 1
V.A. = non
S.A. = non
No Hole
x-int = $\left(1 , 0\right)$ $\left(- 1 , 0\right)$
y-int = $\left(0 , - \frac{1}{4}\right)$

#### Explanation:

. Domain is all real numbers

.When the degree of denominator is equal to the degree of nominator,
H.A.="nominator's leading coefficient"/"denominator's leading coefficient" therefore, $\frac{1}{1}$ = 1 = Horizontal asymptote. .

.For rational functions, V.A. are the undefined points (zeros of the
denominator) of the simplified function.
$\frac{{x}^{2} - 1}{{x}^{2} + 4}$ is true and does not have any undefined points therefore no Vertical Asymptote.

. There are no slant asymptote because there is no V.A.

.There are no holes because there are no common factors in the
nominator and the denominator

. $x$-intercept is a point on the graph where $\left(y = 0\right)$
$\frac{{x}^{2} - 1}{{x}^{2} + 4} = 0$
${x}^{2} - 1 = 0$
x=-0+- sqrt((0^2-4*1(-1))/(2*1) $= + \mathmr{and} - 1$
$x = 1 \mathmr{and} x = - 1$
. $y$-intercept is a point on the graph where $\left(x = 0\right)$
$y = \frac{0 - 1}{0 + 4}$ subtract each number by $0$
$y = \frac{- 1}{4}$
$y = - \frac{1}{4}$