# How do you find vertical, horizontal and oblique asymptotes for f(x) = (7x+1)/(2x-9)?

Aug 6, 2017

Vertical asymptote is at $x = 4.5$ and horizontal asymptote at
$y = 3.5$

#### Explanation:

$f \left(x\right) = \frac{7 x + 1}{2 x - 9}$

For vertical asymptote denominator should be zero ,

$2 x - 9 = 0 \therefore x = \frac{9}{2} = 4.5$. So vertical asymptote is at $x = 4.5$

Here the degree of both denominator and numerator are same

then we have a horizontal asymptote at y = (numerator's leading

coefficient) / (denominator's leading coefficient) i.e at $y = \frac{7}{2}$ or

$y = 3.5$. and no oblique asymptote.

Vertical asymptote is at $x = 4.5$ and

horizontal asymptote at $y = 3.5$

graph{(7x+1)/(2x-9) [-80, 80, -40, 40]}
[Ans]

Aug 6, 2017

$\text{vertical asymptote at } x = \frac{9}{2}$
$\text{horizontal asymptote at } y = \frac{7}{2}$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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 then it is a vertical asymptote.

$\text{solve " 2x-9=0rArrx=9/2" is the asymptote}$

$\text{horizontal asymptotes occur as}$

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

$\text{divide the terms on the numerator/denominator by x}$

$f \left(x\right) = \frac{\frac{7 x}{x} + \frac{1}{x}}{\frac{2 x}{x} - \frac{9}{x}} = \frac{7 + \frac{1}{x}}{2 - \frac{9}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{7 + 0}{2 - 0}$

$\Rightarrow y = \frac{7}{2} \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(7x+1)/(2x-9) [-20, 20, -10, 10]}