Question

Identify the vertical and horizontal asymptotes for f(x) 2x^2 +4/x^2+5x+6. Show your work.

a. x=−2,12x=−2,21​
b. x=−3,2x=−3,2
c. y=0,x=−1y=0,x=−1
d. y=1,x=3y=1,x=3

Answer 1

The function f(x) has vertical asymptotes at x = -2 and x = -3, found by setting the denominator equal to zero, and a horizontal asymptote at y = 2, determined by the ratio of the leading coefficients of the numerator and denominator.

Step-by-step explanation:

The function f(x) = 2x^2 +4 / (x^2 + 5x + 6) has both vertical and horizontal asymptotes. To find them, we need to analyze the behavior of the function as x approaches certain values and as x becomes very large or very small.

First, let's find the vertical asymptotes by determining where the denominator equals zero:

x^2 + 5x + 6 = 0

This is a quadratic equation that can be factored:

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

Therefore, the values that make the denominator zero are x = -2 and x = -3. These values are where the function has vertical asymptotes since the function cannot be defined at these points.

Next, we look for horizontal asymptotes by examining the end-behavior of the function. This is done by comparing the degrees of the polynomial in the numerator and the denominator:

Since the degrees are the same (both quadratics), the horizontal asymptote is found by dividing the leading coefficients of the numerator and the denominator:

y = 2 / 1 = 2

Thus, there is a horizontal asymptote at y = 2.

In conclusion, the vertical asymptotes are x = -2 and x = -3, and the horizontal asymptote is y = 2.