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Phantasy [73]
2 years ago
5

What is the equation of the oblique asymptote? h(x) = x² – 3x - 4/x + 2

Mathematics
1 answer:
NNADVOKAT [17]2 years ago
4 0

Simplifying h(x) gives

h(x) = (x² - 3x - 4) / (x + 2)

h(x) = ((x² + 4x + 4) - 4x - 4 - 3x - 4) / (x + 2)

h(x) = ((x + 2)² - 7x - 8) / (x + 2)

h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)

h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)

h(x) = (x + 2) - 7 - 22/(x + 2)

h(x) = x - 5 - 22/(x + 2)

An oblique asymptote of h(x) is a linear function p(x) = ax + b such that

\displaystyle \lim_{x\to\pm\infty} h(x) - p(x) = 0

In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.

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Given :

Three roots , \sqrt{7},\ -\sqrt{7},\ -4 .

To Find :

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Solution :

Equation of polynomial of 3 zeroes is given by :

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Therefore, the cubic function of given zeroes is x^3+4x^2-7x-28=0.

Hence, this is the required solution.

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Answer:

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