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How do you find vertical, horizontal and oblique asymptotes for # f(x) = (x^2 - 9) / (x - 4)#?

Answer 1

Theodore Amidon

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

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

As you cannot divide by #0#, #x!=4#

The vertical asymptote is #x=4#

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

Let's do a long division

#color(white)(aaaa)##x^2##color(white)(aaaaaa)##-9##color(white)(aaaa)##∣##color(red)(x-4)#

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

#color(white)(aaaaa)##0+4x-9#

#color(white)(aaaaaaa)##+4x-16#

#color(white)(aaaaaaaaa)##+0+7#

Therefore,

#f(x)=(x^2-9)/(x-4)=color(blue)(x+4)+7/(x-4)#

#lim_(x->-oo)(f(x)-(x+4))=lim_(x->-oo)7/x=0^-#

#lim_(x->+oo)(f(x)-(x+4))=lim_(x->+oo)7/x=0^+#

The oblique asymptote is #y=x+4#

graph{(y-(x^2-9)/(x-4))(y-x-4)=0 [-52, 52, -26.03, 26]}

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Answer 2

Caleb Alexander

The vertical asymptote is ( x = 4 ). The horizontal asymptote is ( y = x - 1 ). There are no oblique asymptotes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.