asked 225k views
2 votes
(x^(2)-3x-4)/(3x^(2)-9x) Vertical asymptotes: Holes: none Horizontal asymptote:

asked
User Imjohsep
by
7.8k points

1 Answer

3 votes

Answer: In summary:

Vertical asymptotes: x = 0, x = 3

Holes: none

Horizontal asymptote: y = 1/3

To determine the vertical asymptotes, we need to find the values of x for which the denominator of the expression becomes zero. In this case, the denominator is 3x^2 - 9x.

To find the values of x, we can set the denominator equal to zero and solve for x:

3x^2 - 9x = 0

Factoring out x, we get:

x(3x - 9) = 0

Setting each factor equal to zero, we have:

x = 0 or 3x - 9 = 0

Solving the second equation, we find:

3x = 9

x = 3

Therefore, the vertical asymptotes occur at x = 0 and x = 3.

Next, let's determine if there are any holes in the graph. To do this, we need to check if any factors in the numerator cancel out with factors in the denominator.

In the numerator, we have x^2 - 3x - 4. Factoring this expression, we get:

(x - 4)(x + 1)

In the denominator, we have 3x^2 - 9x. Factoring out 3x, we have:

3x(x - 3)

Comparing the factors, we see that (x - 4) in the numerator does not cancel out with any factors in the denominator. Therefore, there are no holes in the graph.

Finally, let's determine the horizontal asymptote. To do this, we need to compare the degrees of the numerator and denominator.

The degree of the numerator is 2 (highest power of x), and the degree of the denominator is also 2. Since the degrees are the same, we can find the horizontal asymptote by dividing the leading coefficients of the numerator and denominator.

The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is 3. Therefore, the horizontal asymptote is y = 1/3.

Step-by-step explanation:

answered
User Brette
by
8.5k points
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.