How do you find vertical, horizontal and oblique asymptotes for $\frac{3 x^{2} + 2 x - 5}{x - 4}$ $(3x^2+2x-5)/(x-4)$ ?

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Answer 1 · 2017-01-12T08:58:54

The vertical asymptote is $x = 4$ $x=4$
The oblique asymptote is $y = 3 x + 14$ $y=3x+14$
No horizontal asymptote

Let $f (x) = \frac{3 x^{2} + 2 x - 5}{x - 4}$ $f(x)=(3x^2+2x-5)/(x-4)$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{4}$

As we cannot divide by $0$ , $=>$ , $x!=4$

The vertical asymptote is $x=4$

As the degree of the numerator is $>$ than the degree of the denominator, there is an oblique asymptote.

Let's do a long division

$color(white)(aaaa)$ $3x^2+2x-5$ $color(white)(aaaaaa)$ $∣$ $color(blue)(x-4)$

$color(white)(aaaa)$ $3x^2-12x$ $color(white)(aaaaaaaaa)$ $∣$ $color(red)(3x+14)$

$color(white)(aaaaaa)$ $0+14x-5$

$color(white)(aaaaaaaa)$ $+14x-56$

$color(white)(aaaaaaaaaa)$ $+0+51$

Therefore,

$f(x)=(3x+14)+51/(x-4)$

Now, we calculate the limits

$lim_(x->-oo)(f(x)-(3x+14))=lim_(x->-oo)51/x=0^-$

$lim_(x->+oo)(f(x)-(3x+14))=lim_(x->+oo)51/x=0^+$

The oblique asymptote is $y=3x+14$

graph{(y-(3x^2+2x-5)/(x-4))(y-3x-14)=0 [-106, 160.9, -31.5, 102.1]}

How do you find vertical, horizontal and oblique asymptotes for (3x^2+2x-5)/(x-4)3x2+2x−5x−4(3x^2+2x-5)/(x-4)?