How to solve ((x^3 - 8) / (x^2 - 5x + 6))?

Question

I need to show work using long division to find the vertical, horizontal, and oblique asymptotes.

Answer 1

To find horizontal asymptote, first find the degree of the numarator and  the degree of denominator.

Degree of the numarator = 3 and the degree of denominator = 2.

Since the degree of the numerator is one grater than the degree of the denominator, then the function will have slant(oblique) asymptote.

To find the equation of oblique asymptote by long division.

Divide the first term of the dividend by the first term of the divisor x³/x² = x.

So,the first term of the quotient is x. Multiply (x² - 5x + 6) by x and subtract.

Divide the first term of the last row by first term of the divisor 5x²/x² = 5.

So,the second term of the quotient is 5. Multiply (x² - 5x + 6) by 5 and subtract.

The result of the division is (x³ - 8)/(x² - 5x + 6) = (x + 5) + (19x - 38)/(x² - 5x + 6)

Quotient is oblique asymptote .

Oblique asymptote  is y = x + 5.

Vertical asymptote can be found by making denominator = 0 and solve.

x² - 5x + 6 = 0

x² - 3x  - 2x + 6 = 0

x ( x - 3) - 2(x - 3) = 0

(x - 3) (x - 2) = 0

x - 3 = 0 and x - 2 = 0

x = 3 and x = 2

Vertical asymptotes are x = 3 and x = 2.

Comments

Answer is wrong. The vertical asymptote would be 3. If you simplify the equation you would get

x^2 + 2x + 4 / x -3

And from there we can clearly see that the VA is x = 3