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

Question

2534 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-03T05:28:37

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

#y=2x-1# as slanting asymptote

As #f(x)=(2x^2-5x+5)/(x-2)->oo# when #x->2#,

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

Further dividing numerator and denominator by #x#

#f(x)=(2x^2-5x+5)/(x-2)=2x-1+3/(x-2)#

Therefore, when #x->oo#, #f(x)->2x-1#

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

#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