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

Nov 3, 2016

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

$y = 2 x - 1$ as slanting asymptote

Explanation:

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

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

Further dividing numerator and denominator by $x$

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

Therefore, when $x \to \infty$, $f \left(x\right) \to 2 x - 1$

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

$y = 2 x - 5$ 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.