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gregori [183]
3 years ago
5

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

Mathematics
1 answer:
ser-zykov [4K]3 years ago
8 0
So the function we're dealing with is

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

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

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