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Check me please? For the function g(x)=2x^2+1/x+3, what value of x makes this function undefined? I think it's 2. Write the quatiton of the vertical asymptote defined by this value of x. Is it: x=-3?
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A function is undefined when the denominator is 0. So since the denominator is x+3, equate it to 0. x+3=0 x=-3 When x=-3, the denominator becomes zero making the function undefined.
Yes, the vertical asymptote is x=-3.
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