Questions: Find the domain and the x- and y-intercepts of f(x). Then graph the function. f(x) = (x^3 + 5x^2 - 14x) / (x^2 - 2x - 24)

Solution Steps

The function \( f(x) = \frac{x^{3}+5x^{2}-14x}{x^{2}-2x-24} \) is undefined where the denominator is zero. To find these points, solve the equation:

\[ x^{2} - 2x - 24 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -24 \):

\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \] \[ x = \frac{2 \pm \sqrt{4 + 96}}{2} \] \[ x = \frac{2 \pm \sqrt{100}}{2} \] \[ x = \frac{2 \pm 10}{2} \]

Thus, \( x = 6 \) or \( x = -4 \).

The domain of \( f(x) \) is all real numbers except \( x = 6 \) and \( x = -4 \).

The \( x \)-intercepts occur where \( f(x) = 0 \), which is when the numerator is zero:

\[ x^{3} + 5x^{2} - 14x = 0 \]

Factor out an \( x \):

\[ x(x^{2} + 5x - 14) = 0 \]

The solutions are \( x = 0 \) or solving \( x^{2} + 5x - 14 = 0 \) using the quadratic formula:

\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \] \[ x = \frac{-5 \pm \sqrt{25 + 56}}{2} \] \[ x = \frac{-5 \pm \sqrt{81}}{2} \] \[ x = \frac{-5 \pm 9}{2} \]

Thus, \( x = 2 \) or \( x = -7 \).

The \( x \)-intercepts are \( x = 0 \), \( x = 2 \), and \( x = -7 \).

The \( y \)-intercept occurs where \( x = 0 \):

\[ f(0) = \frac{0^{3} + 5 \cdot 0^{2} - 14 \cdot 0}{0^{2} - 2 \cdot 0 - 24} = 0 \]

The \( y \)-intercept is \( y = 0 \).

Final Answer