Question

al alswe to thè question. (2 points) x1 5 5. For the function:2 Part l: a. What value of x makes this function undefined? (2 points) b. Write the equation of the vertical asymptote defined by this value of x. (2 points) Part II: a. Which expression has a larger degree, the numerator or denominator? (2 points) b. Use this information and your choice of long or synthetic division to determine the equation of the oblique asymptote for this function. (4 points)

Answer 1

Part 1:

a) The function is undefined, when the denominator boils down to zero, hence in this case, for the value of x=-3, the function is undefined

b) To find the vertical asymptote, we need to equate the denominator to zero and solve

Part 1: a) The function is undefined, when the denominator boils down to zero, hence in this case, for the value of x = -3, the function is undefined b) To find the vertical asymptote, we need to equate the denominator to zero and solve x + 3 = 0 x = -3 Hence the equation for vertical asymptote will be equal to (x = -3) Part 2: a) The numerator has a x^2 square term(so the degree of numerator is equal to 2) and the denominator has term of x. So the degree of numerator 2 is higher than the denominator b) Using the long division method 2x - 6 x + 3 2x^2 + 1 2 x^2 + 6x -6x + 1 -6x - 18 19 Hence the equation for the oblique asymptote will be (y = 2x - 6)

Hence the equation for vertical asymptote will be equal to (x=-3)

Part 2:

a) The numerator has a x^2 square term(so the degree of numerator is equal to 2) and the denominator has term of x. So the degree of numerator 2 is higher than the denominator

b)

Using the long division method

Hence the equation for the oblique asymptote will be (y=2x-6)