Question: Let \( f(x) = \frac{2x^2 - 3x - 2}{3x^2 - 13x - 10} \) This function has: 1) A y-intercept at…

Let \( f(x) = \frac{2x^2 - 3x - 2}{3x^2 - 13x - 10} \)

This function has:

1. A y-intercept at the point

2. x-intercepts at the point(s)

3. Vertical asymptotes at \( x = \)

4. Horizontal asymptote at \( y = \)

Solution

Step 1: Find the y-intercept by evaluating \( f(0) \). \[ f(0) = \frac{2(0)^2 - 3(0) - 2}{3(0)^2 - 13(0) - 10} = \frac{-2}{-10} = \frac{1}{5} \] The y-intercept is at the point \( (0, \frac{1}{5}) \). Step 2: Find the x-intercepts by setting the numerator equal to zero. \[ 2x^2 - 3x - 2 = 0 \] Solving the quadratic equation: \[ x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \] \[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-2}{4} = -\frac{1}{2} \] The x-intercepts are at the points \( (2, 0) \) and \( \left(-\frac{1}{2}, 0\right) \). Step 3: Find the vertical asymptotes by setting the denominator equal to zero. \[ 3x^2 - 13x - 10 = 0 \] Solving the quadratic equation: \[ x = \frac{13 \pm \sqrt{169 + 120}}{6} = \frac{13 \pm \sqrt{289}}{6} = \frac{13 \pm 17}{6} \] \[ x = \frac{30}{6} = 5 \quad \text{and} \quad x = \frac{-4}{6} = -\frac{2}{3} \] The vertical asymptotes are at \( x = 5 \) and \( x = -\frac{2}{3} \). Step 4: Find the horizontal asymptote by comparing the degrees of the numerator and the denominator. \[ \text{Since the degrees of both the numerator and the denominator are 2, the horizontal asymptote is } y = \frac{2}{3} \] The horizontal asymptote is at \( y = \frac{2}{3} \).