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3 years ago

14

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

1 answer:

eduard3 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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Step-by-step explanation: The representation fog(x) is a representation of composite function, meaning one depends on the other.

In this case, fog(x) means:

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fog(x) = $3(x+\frac{1}{x} )-\frac{1}{x+\frac{1}{x} } -4$

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$fog(x)=\frac{3x^{2}(x^{2}+x)+3(x^{2}+x)-x-4x(x^{2}+x)}{x(x^{2}+x)}$

$fog(x)=\frac{3x^{4}+3x^{3}+3x^{2}+3x-x-4x^{3}+4x^{2}}{x(x^{2}+x)}$

$fog(x)=\frac{3x^{4}-x^{3}-x^{2}+2x}{x(x^{2}+x)}$

This is the function fog(x).

The domain of a function is all the values the independent variable can assume.

For fog(x), denominator can be zero, so:

$x(x^{2}+x) \neq 0$

If x = 0, the function doesn't exist.

$x^{2}+x \neq0$

$x(x+1) \neq0$

$x+1\neq0$

$x\neq-1$

<u>Therefore, the domain of this function is: </u><u>-∞ < -1 or x > 0</u>

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