How do you find vertical, horizontal and oblique asymptotes for #(3x^2+2x-5)/(x-4)#?

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Answer 1 · 2017-01-12T08:58:54

The vertical asymptote is #x=4#
The oblique asymptote is #y=3x+14#
No horizontal asymptote

Let #f(x)=(3x^2+2x-5)/(x-4)#

The domain of #f(x)# is #D_f(x)=RR-{4}#

As we cannot divide by #0#, #=>#, #x!=4#

The vertical asymptote is #x=4#

As the degree of the numerator is #># than the degree of the denominator, there is an oblique asymptote.

Let's do a long division

#color(white)(aaaa)##3x^2+2x-5##color(white)(aaaaaa)##∣##color(blue)(x-4)#

#color(white)(aaaa)##3x^2-12x##color(white)(aaaaaaaaa)##∣##color(red)(3x+14)#

#color(white)(aaaaaa)##0+14x-5#

#color(white)(aaaaaaaa)##+14x-56#

#color(white)(aaaaaaaaaa)##+0+51#

Therefore,

#f(x)=(3x+14)+51/(x-4)#

Now, we calculate the limits

#lim_(x->-oo)(f(x)-(3x+14))=lim_(x->-oo)51/x=0^-#

#lim_(x->+oo)(f(x)-(3x+14))=lim_(x->+oo)51/x=0^+#

The oblique asymptote is #y=3x+14#

graph{(y-(3x^2+2x-5)/(x-4))(y-3x-14)=0 [-106, 160.9, -31.5, 102.1]}