How do you find all the asymptotes for function #f(x) = (2x^2+3x+8)/(x+3)#?

Answer 1

Vertical asymptote #x=3# and slanting or oblique asymptote #y=2x+3#

There are three types of asymptotes.

Vertical asymptotes are indicated by denominator. Here we have #(x+3)# in denominator, so vertical asymptote is given by #x+3=0# or #x=-3#. Observe that in case #x+3# is a factor of numerator, it will cancel out and we will not have an asymptote, rather we will havea hole at #x=-3#. But here #x+3# is not a factor of numerator, we do have a vertical asymptote as
#lim_(x->-3)(2x^2+3x+8)/(x+3)=oo#
Horizontal asymptotes in such cases (algebraic expressions) are there when degree of numerator is equal to that of denominator. Here it is not so we do not a horizontal asymptote. Stll assume numerator as only #3x+8# and then
#lim_(x->oo)(3x+8)/(x+3)=lim_(x->oo)(3+8/x)/(1+3/x)=3/1=3# and we would have #y=3# as horizontal asymptote.
Here we have #(2x^2+3x+8)/(x+3)# and we have
#lim_(x->oo)(2x^2+3x+8)/(x+3)=lim_(x->oo)(2x+3+8/x)/(1+3/x)#
= #2x+3#
Hence we have a slanting or oblique asymptote #y=2x+3#

It is apparent that we have this only when degree of numerator is one more than that of denominator.

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Answer 2

To find the asymptotes of the function ( f(x) = \frac{2x^2+3x+8}{x+3} ), we need to consider both vertical and horizontal asymptotes.

Vertical Asymptotes: Vertical asymptotes occur where the denominator of the function becomes zero but the numerator doesn't. For this function, the vertical asymptote occurs when ( x + 3 = 0 ), so ( x = -3 ) is the vertical asymptote.

Horizontal Asymptote: To find the horizontal asymptote, we need to examine the behavior of the function as ( x ) approaches positive or negative infinity. We can use polynomial division or limits to determine this.

When ( x ) approaches positive or negative infinity, the highest power terms dominate the function. In this case, as ( x ) goes to infinity, ( 2x^2 ) dominates the function. So, the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator, which is ( y = \frac{2}{1} = 2 ).

Therefore, the asymptotes for the function ( f(x) = \frac{2x^2+3x+8}{x+3} ) are:

  • Vertical asymptote: ( x = -3 )
  • Horizontal asymptote: ( y = 2 )
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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.

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.

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.

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.

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