# How do you find the vertical, horizontal and slant asymptotes of:  g(x)= (x+2 )/( 2x^2)?

Jan 1, 2017

The vertical asymptote is $x = 0$
No slant asymptote
The horizontal asymptote is $y = 0$

#### Explanation:

The domain of $g \left(x\right)$ is ${D}_{g} \left(x\right) = \mathbb{R} - \left\{0\right\}$

As you cannot divide by $0$, $x \ne 0$

The vertical asymptote is $x = 0$

The degree of the numerator is $<$ than the degree of the numerator, there is no slant asymptote.

Now, we compute the limits of $g \left(x\right)$ as $x \to \pm \infty$

${\lim}_{x \to - \infty} g \left(x\right) = {\lim}_{x \to - \infty} \frac{x}{2 {x}^{2}} = {\lim}_{x \to - \infty} \frac{1}{2 x} = {0}^{-}$

${\lim}_{x \to + \infty} g \left(x\right) = {\lim}_{x \to + \infty} \frac{x}{2 {x}^{2}} = {\lim}_{x \to + \infty} \frac{1}{2 x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{(x+2)/(2x^2) [-7.9, 7.9, -3.95, 3.95]}

Jan 1, 2017

vertical asymptote at x = 0
horizontal asymptote at y = 0

#### Explanation:

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to x and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : $2 {x}^{2} = 0 \Rightarrow {x}^{2} = 0 \Rightarrow x = 0$

$\Rightarrow x = 0 \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , g \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$g \left(x\right) = \frac{\frac{x}{x} ^ 2 + \frac{2}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2} = \frac{\frac{1}{x} + \frac{2}{x} ^ 2}{2}$

as $x \to \pm \infty , g \left(x\right) \to \frac{0 + 0}{2}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x+2)/(2x^2) [-10, 10, -5, 5]}