How do you identify the zeroes and asymptotesof #f(x) = (4x^2-16)/(x^2-9)#?

1 Answer
Jun 4, 2017

Let me split this up

Explanation:

To find zeros, it means to find x-intercepts, i.e. where #f(x)=0#
In this case, when #f(x)=0#,
#(4x^2-16)/(x^2-9)=0#
Assuming #x^2-9# does not equal zero, we multiply the whole equation by #x^2-9#, giving #4x^2-16=0(x^2-9)#
#therefore 4x^2-16=0#
#x^2=16/4#
#x=+-sqrt4=+-2#
Verifying that #x=+-2# does not result in #x^2-9# to equal 0, (it equals #-5#)
Therefore, #x=+-2# are zeroes for #f(x)#

To find asymptotes, we have to find the x and y values as y and x tend to infinity. Let me split this up into horizontal and vertical asymptotes:

Horizontal asymptote is the y-value as #x->+-oo#, i.e.
#x->+-oo (4x^2-16)/(x^2-9)#
Dividing the numerator and denominator by the variable with the highest power, i.e. #x^2#,
#x->+-oo (4x^2-16)/(x^2-9)=x->+-oo (4-16/x^2)/(1-9/x^2)#
As the variable in the fraction is a squared variable, it means that #x->-oo (4-16/x^2)/(1-9/x^2)=x->+oo (4-16/x^2)/(1-9/x^2)# as the negative sign is removed through the square.
#thereforex->+-oo (4-16/x^2)/(1-9/x^2)=x->oo (4-16/x^2)/(1-9/x^2)=x->oo(4-0)/(1-0)=4#

Therefore, the horizontal asymptote is #y=4#

Vertical asymptotes are x-values that cause the function to carry a not well defined value or indeterminate form, such as #"something"/0#. In other words, to find the vertical asymptote means to solve for #x# when the denominator equals 0, i.e.
#x^2-9=0#
#x^2=9#
#x=+-3#
This means that at #x=+3# and #x=-3#, #f(x)# is undefined, hence the vertical asymptotes for #f(x)# are #x=+-3#