Questions: Find the equations of any vertical asymptotes for the function below f(x)=(x^2+6 x)/(x^3-25 x)

Solution Steps

To find the vertical asymptotes of the function \( f(x) = \frac{x^2 + 6x}{x^3 - 25x} \), we need to determine the values of \( x \) that make the denominator zero, as these are the points where the function is undefined and may have vertical asymptotes. We then check if the numerator is non-zero at these points.

To find the vertical asymptotes of the function \( f(x) = \frac{x^2 + 6x}{x^3 - 25x} \), we first identify the values of \( x \) that make the denominator zero. These are the potential vertical asymptotes.

The denominator is \( x^3 - 25x \). Setting it to zero, we solve: \[ x^3 - 25x = 0 \] Factoring out \( x \), we get: \[ x(x^2 - 25) = 0 \] Further factoring, we have: \[ x(x - 5)(x + 5) = 0 \] Thus, the roots are: \[ x = 0, \quad x = 5, \quad x = -5 \]

Next, we check if the numerator \( x^2 + 6x \) is non-zero at these roots. If the numerator is non-zero at a root of the denominator, then that root is a vertical asymptote.

Evaluating the numerator at \( x = 0 \): \[ 0^2 + 6 \cdot 0 = 0 \] Evaluating the numerator at \( x = 5 \): \[ 5^2 + 6 \cdot 5 = 25 + 30 = 55 \] Evaluating the numerator at \( x = -5 \): \[ (-5)^2 + 6 \cdot (-5) = 25 - 30 = -5 \]

From the evaluations, we see that the numerator is zero at \( x = 0 \), so \( x = 0 \) is not a vertical asymptote. However, the numerator is non-zero at \( x = 5 \) and \( x = -5 \), so these are the vertical asymptotes.

Final Answer