Question

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

Answer 1

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) = $\frac{x^{2}+ 3}{ x^{2} - 9}$ ;

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 
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So; the vertical asymptotes are:  "x = 3" and "x = -3" .
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The horizontal asymptote is:  "y = 2"  .  
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(since:  We have: f(x) = $\frac{x^{2}+ 3}{ x^{2} - 9}$ ;

The "x² / x² " as the highest degree polymonials; both with "implied" coefficients of "1" ; and both raised to the same exponential power of "2".