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maria [59]
3 years ago
6

Solve.

Mathematics
1 answer:
Tema [17]3 years ago
4 0
From 1st equation
d + e = 2
d = 2 - e

From 2nd equation, subtitute d with 2 - e
d - e = 4
2 - e - e = 4
2 - 2e = 4
-2e = 2
e = -1

Find d
d = 2 - e
d = 2 - (-1)
d = 3

The solution is (3,-1)
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The vertical asymptotes are:  "x = 3" and "x = -3" .
__________________________________________
The horizontal asymptote is:  "y = 2"  .  
___________________________________________
Explanation:
___________________________________________

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" ;

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So, what values for "x" exist when "x = 9" ?

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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) = \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".
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