Questions: Find the intercepts and asymptotes. (If an answer does not exist, enter DNE. Enter your asymptotes as a comma-separated list of equations if necessary.) r(x) = (-5x^2 - 50x - 126) / (x^2 + 10x + 25) x-intercept (x, y) = ( ) y-intercept (x, y) = ( ) vertical asymptote(s) horizontal asymptote Sketch a graph of the rational function. State the domain and estimate the range from your graph. (Enter your answers using interval notation.) domain range

Solution Steps

To find the x-intercept(s), set $r(x) = 0$ and solve for $x$.

$0 = \frac{-5x^2 - 50x - 126}{x^2 + 10x + 25}$

$0 = -5x^2 - 50x - 126$

$5x^2 + 50x + 126 = 0$

Using the quadratic formula, we get: $x = \frac{-50 \pm \sqrt{50^2 - 4(5)(126)}}{2(5)} = \frac{-50 \pm \sqrt{2500 - 2520}}{10} = \frac{-50 \pm \sqrt{-20}}{10}$

Since the discriminant is negative, there are no real solutions, so there are no x-intercepts.

To find the y-intercept, set $x=0$ and evaluate $r(0)$.

$r(0) = \frac{-5(0)^2 - 50(0) - 126}{0^2 + 10(0) + 25} = \frac{-126}{25}$

The y-intercept is $(0, -\frac{126}{25})$.

To find the vertical asymptote(s), set the denominator equal to zero and solve for $x$.

$x^2 + 10x + 25 = 0$

$(x+5)^2 = 0$

$x = -5$

So, there is a vertical asymptote at $x=-5$.

Final Answer