How do you find the Vertical, Horizontal, and Oblique Asymptote given $f (x) = \frac{- 10 x + 3}{8 x + 2}$ $f(x)= (-10x+3)/(8x+2)$ ?

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Answer 1 · 2016-11-12T08:23:34

The vertical asymptote is $x = - \frac{1}{4}$ $x=-1/4$
The horizontal asymptote is $y = - \frac{5}{4}$ $y=-5/4$
There are no oblique asymptotes

As we cannot divide by $0$ $0$ , the vertical asymptote is $x = - \frac{1}{4}$ $x=-1/4$

There are no oblique asymptote as the degree of the numeratoris $=$ $=$ the degree of the denominator.

$lim_(x->-oo)f(x)=lim_(x->-oo)(-10x)/(8x)=lim_(x->-oo)-5/4=-5/4$

So, $y=-5/4$ is a horizontal asymptote

graph{(-10x+3)/(8x+2) [-5.03, 4.83, -2.66, 2.272]}

How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (-10x+3)/(8x+2)f(x)=−10x+38x+2f(x)= (-10x+3)/(8x+2)?