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nataly862011 [7]
3 years ago
11

Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

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So for a tangent and a secant, you multiply x^2 (since it’s a tangent you square it) and test that equal to 6(6+12) so now you have x^2=6(6+12). solve for X to get sqrt(108) or 10.4

the equation is — tan^2=outside x whole thing

remember to ADD the entire secant (so add 6 plus 12) don’t multiply them (6x12)

x^2=6(6+12)
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4 0
3 years ago
Read 2 more answers
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