State the vertical asymptote of the rational function.
f(x)=(x-6)(x+6)/x^2-9

Question

State the vertical asymptote of the rational function.
f(x)=(x-6)(x+6)/x^2-9

anonymous · Answer

$${x^2-36}\over{x^2-9}$$if the numerator and denominator are of the same degree, the vertical asymptote is the ratio of $a$ in the numerator to $a$ in the denominator

anonymous · Answer

so is it x=6?

anonymous · Answer

$a$ in the numerator is 1
$a$ in the denominator is 1
$$ratio:{1\over 1}=1$$vertical asymptote: $x=\underline{~~~~}?$

anonymous · Answer

wut isnt it plus or minus 3? assuming its (x^2-9)

anonymous · Answer

x=plus or minus 3?

campbell_st · Answer

factor the denominator and then find the values that make it zero... these are the vertical asymptotes

anonymous · Answer

um, the bottom is kinda important to find asymptotes... I dont know what your talking about

anonymous · Answer

oh wait am i thinking horizontal?.. awk

anonymous · Answer

haha yea im looking for vertical

campbell_st · Answer

so look at the denominator 

$$(x^2 - 9) = (x-3)(x+3)$$

what values make the denominator zero.. they are the asymptotes

anonymous · Answer

so it is +or- 3...right?

anonymous · Answer

lol, this is why we need multiple answerers xD
and yah @adrian111

anonymous · Answer

haha thanks guys!

anonymous · Answer

yea factor the bottom, set the binomials equal to zero and those are your asymptotes...so yes $x=\pm3$ is correct @adrian111 :)

campbell_st · Answer

thats correct... for vertical asymptotes always look for values that make the denominator equal to zero... a zero denominator is undefined... as would occcur with 3 or -3...