Questions: f(x) = (2 x^2 - 2 x - 12) / (x^2 + x - 6) = (2(x - 3)(x + 2)) / ((x + 3)(x - 2)) A) Discontinuities: -3, 2 Holes: None Horz. Asym.: y=0 B) Discontinuities: -3, 2 Holes: None Horz. Asym.: y=2 C) Discontinuities: -2, 3 Holes: None Horz. Asym.: y=1/2 D) Discontinuities: -2, 3 Holes: None Horz. Asym.: y=0

Solution Steps

To solve the given problem, we need to analyze the rational function \( f(x) = \frac{2(x-3)(x+2)}{(x+3)(x-2)} \) to determine its discontinuities, holes, and horizontal asymptotes. Discontinuities occur where the denominator is zero, holes occur where both the numerator and denominator are zero, and horizontal asymptotes are determined by the degrees of the polynomials in the numerator and denominator.

1. Discontinuities: Set the denominator equal to zero and solve for \( x \).
2. Holes: Check if any of the discontinuities also make the numerator zero.
3. Horizontal Asymptote: Compare the degrees of the numerator and denominator.

To find the discontinuities of the function \( f(x) = \frac{2(x-3)(x+2)}{(x+3)(x-2)} \), we set the denominator equal to zero:

\[ (x + 3)(x - 2) = 0 \]

Solving this gives us the discontinuities:

\[ x = -3, \quad x = 2 \]

Next, we check for holes by determining if any of the discontinuities also make the numerator zero. The numerator is \( 2(x-3)(x+2) \). Evaluating the numerator at the discontinuities:

• For \( x = -3 \): \[ 2(-3-3)(-3+2) = 2(-6)(-1) = 12 \quad (\text{not a hole}) \]

• For \( x = 2 \): \[ 2(2-3)(2+2) = 2(-1)(4) = -8 \quad (\text{not a hole}) \]

Since neither discontinuity makes the numerator zero, there are no holes.

To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Both the numerator and denominator are polynomials of degree 2. The horizontal asymptote is given by the ratio of the leading coefficients:

\[ y = \frac{2}{1} = 2 \]

Final Answer