Polynomial

Question

I need to find the domain, simplify, the hole, vertical    Asymptote, horizontal asymptote, x intercept, and y intercept

 

(these are fractions) 

f(x)=  numerator..(x-3)(2x+1)  denominator.. (x+3)^2 (x+2)(x-1)

 

f(x)= numerator.. (x+3)^2(2x+1)

denominator..(x+3)(x+2)(x-1)

Answer 1

For the first f(x)

 

To find the domain, we set the denominator of f(x) equal to zero.  At the same time, this will give us the vertical asymptotes.  Domain is the set of x values where f(x) exists.

 

(x + 3)(x + 3)(x + 2)(x - 1) = 0

 

x = -3   ,   x = -2    ,   x = 1

 

Domain is  (-∞, -3)∪(-3, -2)∪(-2, 1)∪(1, ∞)

 

Vertical asymptotes are   x = -3  ,  x = -2 , and  x = 1

 

 

To find y-intercepts, we evaluate f(x) when x = 0.

 

f(0) = (-3)(1) / (9)(2)(-1)

      = -3/(-18)

      = 1/6

 

The y-intercept is (0, 1/6).

 

 

To find x-intercepts, we set the numerator to zero because f(x) = 0. 

 

(x - 3)(2x + 1) = 0

 

x = 3     and     x = -1/2

 

The x-intercepts are (-1/2 , 0) and (3, 0).

 

 

To find the horizontal asymptotes, we use the degree of the leading terms of the numerator and denominator. 

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is 0. 

If the degree of the numerator and denominator is equal to each other, divide the coefficients of the leading terms. The quotient is the horizontal asymptote.

If the degree of the numerator is greater than the degree of the denominator, their is no steady horizontal asymptote.

 

When we expand the factors in the numerator, the degree of the leading term is 2.  When we expand the factors in the denominator, the degree of the leading term is 4.   2 < 4. 

 

Therefore, the horizontal asymptote is y = 0.

 

Another method to find the horizontal asymptote is to determine the limit of f(x) as x approaches infinity from either the left or right.

 

 

I will let you try the second f(x) on your own.