Questions: Instructions: Identify the vertical horizontal asymptotes of each rational function. x=1 1. f(x) = 2x / (x-1) x-1=0 y = 2 / 1 = 2 3. f(x) = (x^2 - 4x + 3) / (4x + 4) 5. f(x) = 3 / (x^2 + x - 2) vertical asymptote x=1 2. f(x) = -3 S^(-2) / (x+2) 4. f(x) = -3 / (x^2 - 2x - 3) 6. f(x) = (2x + 2) / (x - 3)

Solution Steps

To find the vertical asymptotes of a rational function, set the denominator equal to zero and solve for \( x \). For horizontal asymptotes, compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \). If they are equal, divide the leading coefficients.

To find the vertical asymptote, we set the denominator equal to zero: \[ x - 1 = 0 \implies x = 1 \] Thus, the vertical asymptote is \( x = 1 \).

Next, we determine the horizontal asymptote by comparing the degrees of the numerator and denominator. Both the numerator and denominator are of degree 1. Therefore, we take the ratio of the leading coefficients: \[ y = \frac{2}{1} = 2 \] Thus, the horizontal asymptote is \( y = 2 \).

For this function, we set the denominator equal to zero: \[ 4x + 4 = 0 \implies 4x = -4 \implies x = -1 \] Thus, the vertical asymptote is \( x = -1 \).

The degree of the numerator (2) is greater than the degree of the denominator (1). Therefore, there is no horizontal asymptote.

We find the vertical asymptotes by solving: \[ x^2 + x - 2 = 0 \] Factoring gives: \[ (x - 1)(x + 2) = 0 \implies x = 1 \quad \text{or} \quad x = -2 \] Thus, the vertical asymptotes are \( x = 1 \) and \( x = -2 \).

The degree of the numerator (0) is less than the degree of the denominator (2). Therefore, the horizontal asymptote is: \[ y = 0 \]

Final Answer