SOLUTION: Given p(x) = x^3+5x^2+4x/3(x+1)(x^2−7x+12), find: (a) X-intercept(s) (b) Y-intercept (c) Vertical asymptote(s) (d) Hole(s) (e) End behavior / horizontal asymptote

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Given p(x) = x^3+5x^2+4x/3(x+1)(x^2−7x+12), find: (a) X-intercept(s) (b) Y-intercept (c) Vertical asymptote(s) (d) Hole(s) (e) End behavior / horizontal asymptote      Log On


   



Question 1167485: Given p(x) = x^3+5x^2+4x/3(x+1)(x^2−7x+12),
find:
(a) X-intercept(s)
(b) Y-intercept
(c) Vertical asymptote(s)
(d) Hole(s)
(e) End behavior / horizontal asymptote

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


-intercepts are the points where each is a real zero of the numerator function. The -intercept is the point where and is the numerator function. There is a vertical asymptote at the line for each factor of the denominator function that is not also a factor of the numerator function. There is a hole where ever a linear factor of the denominator has a common linear factor in the numerator.

Horizontal asymptotes:

Case 1: The degree of the numerator polynomial is smaller than the degree of the denominator polynomial -- the line (i.e. the -axis) is the horizontal asymptote.

Case 2: The degree of the numerator polynomial is equal to the degree of the denominator polynomial -- the line where is the lead coefficient of the numerator polynomial and is the lead coefficient of the denominator polynomial.

Case 3: The degree of numerator polynomial is greater than the degree of the denominator polynomial -- there is no horizontal asymptote, rather there is a slant asymptote. The RHS of the equation of the slant asymptote is the quotient obtained from performing polynomial long division excluding any remainder.

John

My calculator said it, I believe it, that settles it


I > Ø