# How do you find the vertical, horizontal and slant asymptotes of: (-2x+7)/(4x-3)?

Aug 24, 2016

vertical asymptote at $x = \frac{3}{4}$
horizontal asymptote at $y = - \frac{1}{2}$

#### Explanation:

Let f(x)$= \frac{- 2 x + 7}{4 x - 3}$

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.

solve: $4 x - 3 = 0 \Rightarrow x = \frac{3}{4} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$f \left(x\right) = \frac{- \frac{2 x}{x} + \frac{7}{x}}{\frac{4 x}{x} - \frac{3}{x}} = \frac{- 2 + \frac{7}{x}}{4 - \frac{3}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{- 2 + 0}{4 - 0}$

$\Rightarrow y = - \frac{1}{2} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1) Hence there are no slant asymptotes.
graph{(-2x+7)/(4x-3) [-10, 10, -5, 5]}