How do you find the vertical, horizontal and slant asymptotes of: #(x^2 - 5)/( x+3)#?

Answer 1

The vertical asymptote is #x=-3#
The slant asymptote is #y=x-3#
There is no horizontal asmptote

The domain of the function is #RR-(-3)# As we cannot divide by 0

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. So we make a long division

#x^2##color(white)(aaaa)##-5##∣##x+3# #x^2+3x##color(white)(aa)##∣##x-3# #0-3x-5# #color(white)(aa)##-3x-9# #color(white)(aaaaa)##0+4#

So we can rewrite the function #(x^2-5)/(x+3)=x-3+4/(x+3)#

So the slant asymptote is #y=x-3#

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Answer 2

Sophie Adams

The vertical asymptote of the function ( \frac{x^2 - 5}{x+3} ) occurs where the denominator equals zero, so x = -3.

There are no horizontal asymptotes for this rational function.

To find the slant asymptote, divide the numerator by the denominator using long division or synthetic division. The quotient represents the equation of the slant asymptote. In this case, the quotient is ( x - 3 ). Therefore, the slant asymptote is ( y = x - 3 ).

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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.