How do you find the asymptotes for $y = \frac{3 x^{2} + x - 4}{2 x^{2} - 5 x}$ $y = (3x^2+x-4) / (2x^2-5x)$ ?

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Answer 1 · 2016-11-25T19:10:56

The vertical asymptotes are $x = 0$ $x=0$ and x=5/2#
No hole, No slant asymptote

The horizontal asymptote is $y = \frac{3}{2}$ $y=3/2$

Let's factorise the denominator

$2 x^{2} - 5 x = x (2 x - 5)$ $2x^2-5x=x(2x-5)$

The domain of $y$ $y$ is $D_y=RR-{0,5/2}$

As you cannot divide by $0$ , $x!=0$ and $x!=5/2$

Therefore, the vertical asymptotes are $x=0$ and x=5/2#

As the degree of the numerator is $=$ than the degree of the denominator, we have no slant asymptote.

For the limits, we take the terms of highest degree

$lim_(x->+-oo)y=lim_(x->+-oo)(3x^2)/(2x^2)=3/2$

The horizontal asymptote is $y=3/2$

graph{(y-((3x^2+x-4)/(2x^2-5x)))(y-3/2)=0 [-14.24, 14.24, -7.12, 7.12]}

How do you find the asymptotes for y = (3x^2+x-4) / (2x^2-5x) y=3x2+x−42x2−5xy = (3x^2+x-4) / (2x^2-5x) ?