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