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identify the vertical asymptotes of f(x)=2/x^2+3x-10
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@hartnn Is that right? ^
i think its -5
yes, but where does that other f(x) come from ? and why that oblique asymptote ? are you copying answers from other site ?
x=2 and x=-5 are correct.
@hartnn Oh, sorry I was answering two problems. My notepad is a little messy sorry. Let me fix that.
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f(x) = 2/(x² + 3x - 10) = 2 / [(x + 5)(x - 2)] Vertical asymptotes are x=2 and x=-5.
thnk you
vertical asymptotes are when denominator goes to 0
@helpineedit Your welcome :)
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