# How do you find vertical, horizontal and oblique asymptotes for y = (2x^2+x+2)/(x+1)?

Dec 8, 2016

Vertical: $\uparrow x = - 1 \downarrow$. Slant: y=2x-1, with slope 2. See illustrative graph.

#### Explanation:

Cross multiplying,

$x y - 2 {x}^{2} - x + y - 2 = x \left(y - 2 x\right) - x + y - 2 = \left(x + 1\right) \left(y - 2 x + 1\right) - 3 = 0$

The form is (ax+by+c)(lx+my+n)=non-zero constant.

So, the given equation represents a hyperbola, with asymptotes

$x + 1 = 0 \mathmr{and} y - 2 x + 1 = 0$

graph{y(x+1)-2x^2-x-2=0 [-40, 40, -20, 20]}

Dec 8, 2016

vertical asymptote at x = - 1
slant asymptote is y = 2x - 1

#### Explanation:

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

solve : $x + 1 = 0 \Rightarrow x = - 1 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

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

y=((2x^2)/x^2+x/x^2+2/x^2)/(x/x^2+1/x^2)=(2+1/x+2/x^2)/(1/x+1/x^2

as $x \to \pm \infty , y \to \frac{2 + 0 + 0}{0 + 0} = \frac{2}{0}$

This is undefined hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here (numerator-degree 2 , denominator- degree 1 ) Hence there is an oblique asymptote.

Using $\textcolor{b l u e}{\text{polynomial division}}$

$y = 2 x - 1 + \frac{3}{x + 1}$

as $x \to \pm \infty , y \to 2 x - 1 + 0$

$\Rightarrow y = 2 x - 1 \text{ is the asymptote}$
graph{(2x^2+x+2)/(x+1) [-46.24, 46.24, -23.12, 23.12]}