How do you find vertical, horizontal and oblique asymptotes for #y = (2x^2+x+2)/(x+1)#?

2 Answers
Dec 8, 2016

Answer:

Vertical: #uarrx = -1darr#. Slant: y=2x-1, with slope 2. See illustrative graph.

Explanation:

Cross multiplying,

#xy-2x^2-x+y-2=x(y-2x)-x+y-2=(x+1)(y-2x+1)-3=0#

The form is (ax+by+c)(lx+my+n)=non-zero constant.

So, the given equation represents a hyperbola, with asymptotes

#x+1 = 0 and y-2x+1=0#

graph{y(x+1)-2x^2-x-2=0 [-40, 40, -20, 20]}

Dec 8, 2016

Answer:

vertical asymptote at x = - 1
slant asymptote is y = 2x - 1

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value the it is a vertical asymptote.

solve : #x+1=0rArrx=-1" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" ( a constant)"#

divide all terms on numerator/denominator by the highest power of x, that is #x^2#

#y=((2x^2)/x^2+x/x^2+2/x^2)/(x/x^2+1/x^2)=(2+1/x+2/x^2)/(1/x+1/x^2#

as #xto+-oo,yto(2+0+0)/(0+0)=2/0#

This is undefined hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here (numerator-degree 2 , denominator- degree 1 ) Hence there is an oblique asymptote.

Using #color(blue)"polynomial division"#

#y=2x-1+3/(x+1)#

as #xto+-oo,yto2x-1+0#

#rArry=2x-1" is the asymptote"#
graph{(2x^2+x+2)/(x+1) [-46.24, 46.24, -23.12, 23.12]}