Identify the vertical asymptotes of f(x)=x+6 over x^2-9x+18

QUESTION POSTED AT 01/06/2020 - 04:44 PM

Answered by admin AT 01/06/2020 - 04:44 PM

\bf \cfrac{x+6}{x^2-9x+18}
\\\\\\
x^2-9x+18=\underline{0}\implies (x-6)(x-3)=0\implies x=
\begin{cases}
6\\
3
\end{cases}

vertical asymptotes occur, at zeros of the denominator, so long the value doesn't make the numerator zero.

so, in this case are x = 6 and x = 3.
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