How do you find the asymptotes for $f (x) = \frac{2 x^{2} - 5 x + 6}{x + 2}$ $f(x) =(2x^2-5x+6)/(x+2)$ ?

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Answer 1 · 2016-12-16T07:57:37

The vertical asymptote is $x = - 2$ $x=-2$
The slant asymptote is $y = 2 x - 9$ $y=2x-9$
No horizontal asymptote

As we cannot divide by $0$ $0$ , $x \neq - 2$ $x!=-2$

The degree of the numerator is $>$ $>$ than the degree of the denominator, we expect a slant asymptote.

Let's do a long division

$a a a a$ $color(white)(aaaa)$ $2 x^{2} - 5 x + 6$ $2x^2-5x+6$ $∣$ $∣$ $x + 2$ $x+2$

$a a a a$ $color(white)(aaaa)$ $2 x^{2} + 4 x$ $2x^2+4x$ $a a a a$ $color(white)(aaaa)$ $∣$ $∣$ $2 x - 9$ $2x-9$

$a a a a a$ $color(white)(aaaaa)$ $0 - 9 x + 6$ $0-9x+6$

$a a a a a a a$ $color(white)(aaaaaaa)$ $- 9 x - 18$ $-9x-18$

$a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaa)$ $+ 24$ $+24$

So,

$\frac{2 x^{2} - 5 x + 6}{x + 2} = 2 x - 9 + \frac{24}{x + 2}$ $(2x^2-5x+6)/(x+2)=2x-9+24/(x+2)$

The slant asymptote is $y = 2 x - 9$ $y=2x-9$

For the horizontal asymptotes, we calculate the limita as $x_{>} \infty$ $x_>oo$

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

There is no horizontal asymptote.

graph{(y-(2x^2-5x+6)/(x+2))(y-2x+9)=0 [-93.7, 93.9, -46.8, 46.9]}

How do you find the asymptotes for f(x) =(2x^2-5x+6)/(x+2)f(x)=2x2−5x+6x+2f(x) =(2x^2-5x+6)/(x+2)?