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2 answers:
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The vertical asymptotes are: "x = 3" and "x = -3" .
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The horizontal asymptote is: "y = 2" .
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Explanation:
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f(x) =
;
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 <span>equation ; and</span> to solve for "x" ;
√(x²) = √9 ;
x = ± 3
<span>__________________________________________
So; the vertical asymptotes are: "x = 3" and "x = -3" .
__________________________________________
The horizontal asymptote is: "y = 2" .
__________________________________________
(since: We have: </span>f(x) = ;
The "x² / x² " as the highest degree polymonials; both with "implied" coefficients of "1" ; and both raised to the same exponential power of "2".
__________________________________________
The horizontal asymptote is: "y = 2" .
___________________________________________
Explanation:
___________________________________________
f(x) =
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 <span>equation ; and</span> to solve for "x" ;
√(x²) = √9 ;
x = ± 3
<span>__________________________________________
So; the vertical asymptotes are: "x = 3" and "x = -3" .
__________________________________________
The horizontal asymptote is: "y = 2" .
__________________________________________
(since: We have: </span>f(x) =
The "x² / x² " as the highest degree polymonials; both with "implied" coefficients of "1" ; and both raised to the same exponential power of "2".