SOLUTION: 5) Find the equations of the horizontal and vertical asymptotes for the following. Type none if the function does not have an asymptote. a) f (x)= 2x+3/x+2 Answer

Algebra ->  Functions -> SOLUTION: 5) Find the equations of the horizontal and vertical asymptotes for the following. Type none if the function does not have an asymptote. a) f (x)= 2x+3/x+2 Answer      Log On


   



Question 125252:
5) Find the equations of the horizontal and vertical asymptotes for the following. Type none if the function does not have an asymptote.
a) f (x)= 2x+3/x+2


Answer:
Horizontal:
Vertical:
b) g (x) = 5x/x^+1
Answer:
Horizontal:
Vertical:


Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)


y=%282x%2B3%29%2F%28x%2B2%29%29 Start with the given function



Looking at the numerator 2x%2B3, we can see that the degree is 1 since the highest exponent of the numerator is 1. For the denominator x%2B2, we can see that the degree is 1 since the highest exponent of the denominator is 1.


Horizontal Asymptote:
Since the degree of the numerator and the denominator are the same, we can find the horizontal asymptote using this procedure:

To find the horizontal aysmptote, first we need to find the leading coefficients of the numerator and the denominator.

Looking at the numerator 2x%2B3, the leading coefficient is 2

Looking at the denominator x%2B2, the leading coefficient is 1

So the horizontal aysmptote is the ratio of the leading coefficients. In other words, simply divide 2 by 1 to get %282%29%2F%281%29=2


So the horizontal asymptote is y=2





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Vertical Asymptote:
To find the vertical aysmptote, just set the denominator equal to zero and solve for x

x%2B2=0 Set the denominator equal to zero


x=0-2Subtract 2 from both sides


x=-2 Combine like terms on the right side


So the vertical asymptote is x=-2


Notice if we graph y=%282x%2B3%29%2F%28x%2B2%29, we can visually verify our answers:

Graph of y=%282x%2B3%29%2F%28x%2B2%29%29 with the horizontal asymptote y=2 (blue line) and the vertical asymptote x=-2 (green line)







b)



y=%285x%29%2F%28x%5E2%2B1%29%29 Start with the given function



Looking at the numerator 5x, we can see that the degree is 1 since the highest exponent of the numerator is 1. For the denominator x%5E2%2B1, we can see that the degree is 2 since the highest exponent of the denominator is 2.


Horizontal Asymptote:

Since the degree of the numerator (which is 1) is less than the degree of the denominator (which is 2), the horizontal asymptote is always y=0

So the horizontal asymptote is y=0



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Vertical Asymptote:
To find the vertical aysmptote, just set the denominator equal to zero and solve for x

x%5E2%2B1=0 Set the denominator equal to zero


x%5E2=0-1Subtract 1 from both sides


x%5E2=-1 Combine like terms on the right side


x=0%2B-sqrt%28-1%29 Take the square root of both sides

Since you cannot take the square root of a negative number, the answer is not a real number. So in this case, there are no vertical asymptotes.




Notice if we graph y=%285x%29%2F%28x%5E2%2B1%29, we can visually verify our answers:

Graph of y=%285x%29%2F%28x%5E2%2B1%29%29 with the horizontal asymptote y=0 (blue line) . Notice how there are no vertical asymptotes