determine the horizontal asymptote of the graph of the funct

determine the horizontal asymptote of the graph of the function:

a. y=1

b. y=4/3

c. y=0

d. y=7/4

Solution

f(x) = (4x³ - 2x² + 3x + 7)/(4 - 8x + 3x³ + x²)

==> f(x) = (4x³ - 2x² + 3x + 7)/(3x³ + x² - 8x -4)

leading coefficient in the numerator = 4 , leading coefficiet in the denominator = 3

Equation of harizontal asymptote y = (leading coefficient in the numerator)/( leading coefficiet in the denominator)

==> y = 4/3

Hence y = 4/3 is the equation of horizontal asymptote. (option b)

Method 2:

Equation of horizontal asymptote y = lim_{[x -> ]} f(x)

==> y = lim_{[x -> ]} (4x³ - 2x² + 3x + 7)/(4 - 8x + 3x³ + x²)

==> y = lim_{[x -> ]} (x³(4 - 2/x + 3/x² + 7/x³))/(x³( 4/x³ - 8/x² + 3 + 1/x)

==> y = lim_{[x -> ]} (4 - 2/x + 3/x² + 7/x³)/( 4/x³ - 8/x² + 3 + 1/x)

==> y = (4 - 0 + 0 + 0)/(0 - 0 + 3 + 0)

==> y = 4/3 is equation of horizontal asymptote (Option B)