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2 years ago

Find the horizontal for oblique asymptote of f(x)=-3x^2+7x+1/x-2

1 answer:

eduard2 years ago

3 0

For greater clarity, please enclose the divisor (x-2) inside parentheses. Thx.

Divide x-2 into 3x^2 + 7x + 1:

3x + 13
__________________
x-2 / 3x^2 + 7x + 1
3x^2 - 6x
----------------
13x + 1
13x - 26
----------------
-27 (remainder)
-27
The quotient here is 3x + 13 - ------
x-2

The slant asymptote is the line y = 3x + 13. Note that -27 / (x-2) goes to zero as x grows large in either direction (positive or neg).

Note that you are looking for the "oblique" or "slant" asymptote here; there is no "horizontal" involved.

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We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

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Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

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If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

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