How do you graph #y=x^3/(x^2+7)# using asymptotes, intercepts, end behavior?

1 Answer
Nov 20, 2016

Answer:

See explanation below

Explanation:

The domain of y is #D_y=RR#

The denominator is #(x^2+7)>0#, so we don't have a vertical asymptote.

As the degree of the numerator is #># the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##x^3##color(white)(aaaaaaaa)##∣##x^2+7#

#color(white)(aaaa)##x^3+7x##color(white)(aaaa)##∣##x#

#color(white)(aaaaa)##0-7x#

So, #x^3/(x^2+7)=x-(7x)/(x^2+7)#

Therefore, #y=x# is an oblique asymptote

For the limits, we take the terms of highest degree in the numerator and the denominator

#lim_(x->+-oo)y=lim_(x->+-oo)x^3/x^2=lim_(x->+-oo)x=+-oo#

For the intercepts, when #x=0#, #=># #y=0#

graph{(y-x^3/(x^2+7))(y-x)=0 [-7.9, 7.904, -3.956, 3.946]}