# How do you find the asymptotes for f(x)=[(5x+3)/(2x-3)]+1?

Jul 19, 2016

vertical asymptote x $= \frac{3}{2}$
horizontal asymptote $y = \frac{7}{2}$

#### Explanation:

The first step is to express f(x) as a single fraction with common denominator (2x - 3)

$f \left(x\right) = \frac{5 x + 3}{2 x - 3} + \frac{2 x - 3}{2 x - 3} = \frac{7 x}{2 x - 3}$

The denominator of f(x) cannot be zero as this is 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 : $2 x - 3 = 0 \Rightarrow x = \frac{3}{2} \text{ is the asymptote}$

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

$\frac{\frac{7 x}{x}}{\frac{2 x}{x} - \frac{3}{x}} = \frac{7}{2 - \frac{3}{x}}$

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

$\Rightarrow y = \frac{7}{2} \text{ is the asymptote}$
graph{(7x)/(2x-3) [-20, 20, -10, 10]}