Identify the vertical asymptotes of f(x)=x+6 over x^2-9x+18
1 answer:
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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Step-by-step explanation:
2u + 4u - 3u = 18
6u - 3u = 18
3u = 18
3u/3 = 18/3
u = 6
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Since it’s half a number you do
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