How do you find the vertical, horizontal and slant asymptotes of: $y = \frac{x^{2} + 7}{5 x - 4 x^{2}}$ $y = (x^2 + 7)/( 5x - 4x^2)$ ?

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Answer 1 · 2017-03-09T12:56:09

The vertical asymptotes are $x = 0$ $x=0$ and $x = \frac{5}{4}$ $x=5/4$
The horizontal asymptote is $y = - \frac{1}{4}$ $y=-1/4$
No slant asymptote

We cannot divide by $0$ $0$ , so the denominator cannot $= 0$ $=0$

So,

$5 x - 4 x^{2} \neq 0$ $5x-4x^2!=0$

$x (5 - 4 x) \neq 0$ $x(5-4x)!=0$

Therefore, $x \neq 0$ $x!=0$ and $x \neq \frac{5}{4}$ $x!=5/4$

The vertical asymptotes are $x = 0$ $x=0$ and $x = \frac{5}{4}$ $x=5/4$

The degree of the the numerator $=$ $=$ the degree of the denominator, there is no slant asymptote

$lim_(x->+-oo)y=lim_(x->+-oo)-x^2/(4x^2)=-1/4$

The horizontal asymptote is $y=-1/4$

graph{(y-(x^2+7)/(5x-4x^2))(y+1/4)(y-1000(x-5/4))=0 [-13.42, 14.3, -4.98, 8.89]}

How do you find the vertical, horizontal and slant asymptotes of: y = (x^2 + 7)/( 5x - 4x^2)y=x2+75x−4x2y = (x^2 + 7)/( 5x - 4x^2)?