Questions: Find the x-and y-intercepts, and the equations of the vertical, and horizontal asymptotes. If there are multiple of a specific intercept or asymptote, list them in order from least to greatest. If there is none, enter "DNE". After you complete your analysis, provide a graph of your rational function on paper to turn in with your exam. Be sure to include your asymptotes. f(x)=(3 x^2-13 x-10)/(2 x^2+11 x+5) x-intercepts: , 0) y-intercept (0, Horizontal asymptote at y= Vertical asymptote at x= , Now create a graph of your rational function on paper and turn in your drawing at the conclusion of your exam.

Solution Steps

To find the $x$-intercepts, set the numerator equal to zero and solve for $x$: \[ 3x^2 - 13x - 10 = 0 \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -13\), and \(c = -10\): \[ x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3} \] \[ x = \frac{13 \pm \sqrt{169 + 120}}{6} \] \[ x = \frac{13 \pm \sqrt{289}}{6} \] \[ x = \frac{13 \pm 17}{6} \] The solutions are: \[ x = \frac{30}{6} = 5, \quad x = \frac{-4}{6} = -\frac{2}{3} \] Thus, the $x$-intercepts are \((-2/3, 0)\) and \((5, 0)\).

To find the $y$-intercept, evaluate \(f(0)\): \[ f(0) = \frac{3(0)^2 - 13(0) - 10}{2(0)^2 + 11(0) + 5} = \frac{-10}{5} = -2 \] Thus, the $y$-intercept is \((0, -2)\).

For horizontal asymptotes, compare the degrees of the numerator and the denominator. Both are degree 2, so the horizontal asymptote is: \[ y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{3}{2} \] Thus, the horizontal asymptote is \(y = \frac{3}{2}\).

To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\): \[ 2x^2 + 11x + 5 = 0 \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 11\), and \(c = 5\): \[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \] \[ x = \frac{-11 \pm \sqrt{121 - 40}}{4} \] \[ x = \frac{-11 \pm \sqrt{81}}{4} \] \[ x = \frac{-11 \pm 9}{4} \] The solutions are: \[ x = \frac{-2}{4} = -\frac{1}{2}, \quad x = \frac{-20}{4} = -5 \] Thus, the vertical asymptotes are \(x = -5\) and \(x = -\frac{1}{2}\).

Final Answer