How do you identify all horizontal and slant asymptote for $f (x) = \frac{2 x^{2} - 5 x + 5}{x - 2}$ $f(x)=(2x^2-5x+5)/(x-2)$ ?

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Answer 1 · 2016-11-03T05:28:37

We have $x = 2$ $x=2$ as a vertical asymptote and

$y = 2 x - 1$ $y=2x-1$ as slanting asymptote

As $f (x) = \frac{2 x^{2} - 5 x + 5}{x - 2} \to \infty$ $f(x)=(2x^2-5x+5)/(x-2)->oo$ when $x \to 2$ $x->2$ ,

we have a vertical asymptote at $x = 2$ $x=2$ .

Further dividing numerator and denominator by $x$ $x$

$f (x) = \frac{2 x^{2} - 5 x + 5}{x - 2} = 2 x - 1 + \frac{3}{x - 2}$ $f(x)=(2x^2-5x+5)/(x-2)=2x-1+3/(x-2)$

Therefore, when $x \to \infty$ $x->oo$ , $f (x) \to 2 x - 1$ $f(x)->2x-1$

Hence, we have $x = 2$ $x=2$ as a vertical asymptote and

$y = 2 x - 5$ $y=2x-5$ as slanting asymptote.
graph{(y-(2x^2-5x+5)/(x-2))(y-2x+1)=0 [-19.32, 20.68, -6.64, 13.36]}

Hi - This is Amory. Sorry but your solution is incorrect. This is the graph of the function and your SA.

enter image source here

How do you identify all horizontal and slant asymptote for f(x)=(2x^2-5x+5)/(x-2)f(x)=2x2−5x+5x−2f(x)=(2x^2-5x+5)/(x-2)?