jdepaolo15 jdepaolo15
  • 25-06-2018
  • Mathematics
contestada

dentify the vertical asymptotes of f(x) = quantity x plus 6 over quantity x squared minus 9x plus 18.

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Jrdan
Jrdan Jrdan
  • 25-06-2018
So the function we're dealing with is

f(x) = [tex] \frac{x+6}{x^{2}-9x+18 } [/tex]

Factorising we get, [tex]x^{2} -9x+18=0[/tex]
[tex](x-3)(x-6)=0[/tex]

Therefore, the vertical asymptotes are at x = 3 and x = 6 
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