How do you find the asymptotes for $(3x^2+x-4) / (2x^2-5x)$ ?

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Answer 1 · 2016-02-01T11:24:43

Horizontal asymptote:
$y=3/2$

Vertical asymptote:
$x=0$
$x=5/2$

Here is a graph of the expression in question.
graph{(3x^2 + x - 4)/(2x^2 - 5x) [-10, 10, -5, 5]}
You can try to simplify the expression first.

$(3x^2 + x - 4)/(2x^2 - 5x) = 3/2 + frac{17x - 8}{4x^2 - 10x}$

When $x$ is very large (or small), observe that the second term tends to zero. Therefore,

$lim_{x->-oo} frac{3x^2 + x - 4}{2x^2 - 5x} = 3/2$ and

$lim_{x->oo} frac{3x^2 + x - 4}{2x^2 - 5x} = 3/2$ .

Hence, there is a horizontal asymptote with equation $y=3/2$ .

Also note that the expression is not defined for $x=5/2$ and $x=0$ , as they result in division by zero.

Hence, there are 2 vertical asymptotes, with equation $x=5/2$ and $x=0$ .

How do you find the asymptotes for (3x^2+x-4) / (2x^2-5x)(3x^2+x-4) / (2x^2-5x)?