How would you find the equations of both the horizontal and vertical asymptotes for the following functions: #y=(x)/(x-3)#? #y=(x+4)/(x^2-1)#? #y=(x+4)/(x^2-1)#? #y=(x^2-2x+1)/(x^2-3x-4)#? #y=(x^2-9)/(x^2-3x^2-18x)#?

In addition, how would you find the coordinates of any holes for these problems?

3 Answers
Aug 8, 2016

Answer:

vertical asymptote at x = 3
horizontal asymptote at y = 1

Explanation:

For y #=x/(x-3)#

The denominator of y cannot be zero as this is undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: x - 3 = 0 #rArrx=3" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" (a constant)"#

divide terms on numerator/denominator by x

#(x/x)/(x/x-3/x)=1/(1-3/x)#

as #xto+-oo,yto1/(1-0)#

#rArry=1" is the asymptote"#
graph{(x)/(x-3) [-10, 10, -5, 5]}

Aug 8, 2016

Answer:

vertical asymptotes at x = ± 1
horizontal asymptote at y = 0

Explanation:

For #y=(x+4)/(x^2-1)#

The explanation for vertical/horizontal asymptotes is as described above in previous question.

solve: #x^2-1=0rArr(x-1)(x+1)=0rArrx=±1#

#rArrx=-1" and " x=1" are the asymptotes"#

To obtain horizontal asymptote divide terms on numerator/denominator by the highest power of x, that is #x^2#

#(x/x^2+4/x^2)/(x^2/x^2-1/x^2)=(1/x+4/x^2)/(1-1/x^2)#

as #xto+-oo,yto(0+0)/(1-0)#

#rArry=0" is the asymptote"#
graph{(x+4)/(x^2-1) [-10, 10, -5, 5]}

Aug 11, 2016

Answer:

For: #y = (x^2 - 2x + 1)/(x^2 - 3x - 4)#

Explanation:

Start by factoring both the numerator and the denominator:

#y =((x - 1)(x - 1))/((x - 4)(x + 1)#

Nothing can get canceled out, so we don't have any holes. Now for asymptotes. The vertical asymptotes can be found by setting the denominator to #0# and solving for x.

#x^2 - 3x - 4 = 0#

#(x - 4)(x + 1) = 0#

#x = 4 and -1#

So, the equations of the vertical asymptotes will be #x = 4# and #x = -1#.

For the horizontal asymptotes, we must examine the degree of the denominator relative to that of the numerator.

•The denominator is of degree #2#.
•The numerator is of degree #2#.

When the degrees are equal, the horizontal asymptote occurs at the ratio between the coefficients of the highest degree in the numerator and in the denominator.

Hence, #"asymptote"_"horizontal" = (1 xx x^2)/(1 xx x^2) = 1/1 = 1#

So, there will be a horizontal asymptote at #y = 1#.

In summary:

The graph of #y = (x^2 - 2x + 1)/(x^2 - 3x - 4)# will have vertical asymptotes at #x = 4# and #x = -1#. It will have a horizontal asymptote at #y = 1#. The function will have no holes.

Bonus:

Here is the graph of the function. As you can see, the asymptotes are just where algebra showed them to be.

enter image source here

Hopefully this helps!