How do you find the slant asymptote of #f(x) =( x^2 + 3x - 3) / (x+4)#?

2 Answers
Jan 15, 2017

Answer:

The slant asymptote is #y=x-1#

Explanation:

Let's do the long division

#color(white)(aaaa)##x^2+3x-3##color(white)(aaaa)##∣##color(red)(x+4)#

#color(white)(aaaa)##x^2+4x##color(white)(aaaaaaaa)##∣##color(blue)(x-1)#

#color(white)(aaaaa)##0-x-3#

#color(white)(aaaaaaa)##-x-4#

#color(white)(aaaaaaaaa)##0+1#

Therefore,

#(x^2+3x-3)/(x+4)=x-1+1/(x+4)#

Let #f(x)=(x^2+3x-3)/(x+4)#

So,

#f(x)=x-1+1/(x+4)#

#lim_(x->-oo)f(x)-(x-1)=lim_(x->-oo)1/x=0^-#

#lim_(x->+oo)f(x)-(x-1)=lim_(x->+oo)1/x=0^+#

The slant asymptote is #y=x-1#

graph{(y-(x^2+3x-3)/(x+4))(y-x+1)=0 [-10, 10, -5, 5]}

Jan 15, 2017

Answer:

Slant asymptote: #y = x-1#. See the asymptotes-inclusive graph.

Explanation:

By division,

#y = f(x)= x-1+1/(x+4)#

Rearranged,

#(y-x+1)(x+4)=1#, revealing that the graph is a hyperbola with

asymptotes

#(y-x+1)(x+4)=0#.

Separately,

y-x+1=0 gives a slant asymptote and

x+4=0 gives a vertical asymptote.

graph{((y-x+1)(x+4)-1)(y-x+1)(x+0.000001y+4)=0 [-23, 23.02, -12.7, 12.6]}