Given fx x2 5x 4x3 find its vertical asymptote slant asymp

Given f(x) = x^2 + 5x +4/x-3, find its vertical asymptote slant asymptote [show how you found its slant asymptote by using either long division or synthetic division]

Solution

For a function , vertical asymptote is the point at which the function is not defined.

Or in simple words when the denominator becomes zero.

As the denominator is x-3, consider x -3 = 0 .

Hence x = 3

Hence at x = 3 , the denominator will become zero and the function will become undefined.

Hence vertical asymptote is x = 3.

As the degree of the numerator is 1 + degree of the denominator , the horizontal asymptote is the slant asymptote. of the type y =mx + b.

For a rational function , the slant asymptote is the quotient of the polynomial division.

Divide : (x² + 5x +4) / (x-3) = x + (8x+4) /(x-3) = x + 8 + 28 / (x-3)

Hence : quotient is x + 8 and remainder is 28.

Hence slant asymptote is x + 8.