Question

Use long divsion to find oblique asymptote of f(x)=2x³/3(x-2)²

a. y= 2/3x-4/3
b. y= 2/3x+4/3
c. y= 2/3x
d. y= 2/3x²-4/3x+8/3

Answer 1

The oblique asymptote of
`\(f(x) = (2x^3)/(3(x-2)^2)\) is given by \(y = (2)/(3)x - (4)/(3)\).`

Step-by-step explanation:

To find the oblique asymptote, we perform long division between
`\(2x^3\) and \(3(x-2)^2\). The result is \((2)/(3)x - (4)/(3)\)`, which represents the oblique asymptote.

Long division involves dividing the numerator by the denominator. In this case, we divide
`\(2x^3\) by \(3(x-2)^2\). The quotient is \((2)/(3)x\)` and the remainder is
`\(-(4)/(3)\).` The oblique asymptote is determined by the quotient, and the remainder indicates the vertical shift.

The oblique asymptote
`\(y = (2)/(3)x - (4)/(3)\) implies that as \(x\)`approaches positive or negative infinity, the function \
`(f(x)\)` approaches the line
`\((2)/(3)x - (4)/(3)\).` This occurs when the degree of the numerator is one greater than the degree of the denominator in the rational function. In this case, the oblique asymptote provides insight into the long-term behavior of the function.