What is the equation of the oblique asymptote $f (x) = \frac{x^{2} - 5 x + 6}{x - 4}$ $f(x) = (x^2-5x+6)/(x-4)$ ?

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Answer 1 · 2018-07-09T08:51:06

The equation of the asymptote is $y = x - 1$ $y=x-1$

Perform a long divoision

$a a a a$ $color(white)(aaaa)$ $x^{2} - 5 x + 6$ $x^2-5x+6$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x - 4$ $x-4$

$a a a a$ $color(white)(aaaa)$ $x^{2} - 4 x$ $x^2-4x$ $a a a a a a a$ $color(white)(aaaaaaa)$ $∣$ $|$ $x - 1$ $x-1$

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

$a a a a a a a$ $color(white)(aaaaaaa)$ $- x + 4$ $-x+4$

$a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $0 + 2$ $0+2$

Therefore,

$f (x) = \frac{x^{2} - 5 x + 6}{x - 4} = (x - 1) + \frac{2}{x - 4}$ $f(x)=(x^2-5x+6)/(x-4)=(x-1)+2/(x-4)$

To determine the slant asymptotes, determine the limits

$lim_(x->+oo)( f(x)-(x-1))=lim_(x->+oo)2/(x-4)=0^+$

The curve is above the asymptote

$lim_(x->-oo)( f(x)-(x-1))=lim_(x->-oo)2/(x-4)=0^-$

The curve is below the asymptote

The equation of the asymptote is $y=x-1$

graph{(y-(x^2-5x+6)/(x-4))(y-x+1)=0 [-23.34, 22.27, -8.58, 14.23]}

What is the equation of the oblique asymptote f(x) = (x^2-5x+6)/(x-4)f(x)=x2−5x+6x−4 f(x) = (x^2-5x+6)/(x-4)?