For the given function, find the vertical and horizontal asymptote(s) (if there are any). f(x) = the quantity x squared plus three divided by the quantity x squared minus nine
The vertical asymptotes are: "x = 3" and "x = -3" . __________________________________________ The horizontal asymptote is: "y = 2" . ___________________________________________ Explanation: ___________________________________________
f(x) = [tex] \frac{x^{2}+ 3}{ x^{2} - 9}[/tex] ;
We know that "(x² − 9) ≠ 0 ; since we cannot divide by "0" ; so the "denominator" in the fraction cannot be "0" ;
since: 9 − 9 = 0 ; "x² " cannot equal 9.
So, what values for "x" exist when "x = 9" ?
x² = 9 ;
Take the square root of EACH SIDE of the equation ; to isolate "x" on one side of the equation ; and to solve for "x" ;
√(x²) = √9 ;
x = ± 3 __________________________________________ So; the vertical asymptotes are: "x = 3" and "x = -3" . __________________________________________ The horizontal asymptote is: "y = 2" . __________________________________________ (since: We have: f(x) = [tex] \frac{x^{2}+ 3}{ x^{2} - 9}[/tex] ;
The "x² / x² " as the highest degree polymonials; both with "implied" coefficients of "1" ; and both raised to the same exponential power of "2".
