1. **Problem Statement:** Consider the function $$f(x) = \frac{1}{x^2 + x - 6}$$. We need to find: - Domain and range - Intercepts - Equations of any asymptotes 2. **Step 1: Factor the denominator** The denominator is a quadratic: $$x^2 + x - 6$$. Factor it: $$x^2 + x - 6 = (x + 3)(x - 2)$$. 3. **Step 2: Domain** The function is undefined where the denominator is zero: $$x + 3 = 0 \Rightarrow x = -3$$ $$x - 2 = 0 \Rightarrow x = 2$$ So, the domain is all real numbers except $$x = -3$$ and $$x = 2$$: $$\text{Domain} = (-\infty, -3) \cup (-3, 2) \cup (2, \infty)$$. 4. **Step 3: Intercepts** - **y-intercept:** Set $$x=0$$: $$f(0) = \frac{1}{0^2 + 0 - 6} = \frac{1}{-6} = -\frac{1}{6}$$ So, y-intercept is $$\left(0, -\frac{1}{6}\right)$$. - **x-intercept:** Set $$f(x) = 0$$: $$\frac{1}{x^2 + x - 6} = 0$$ This fraction equals zero only if numerator is zero, but numerator is 1, so no x-intercepts. 5. **Step 4: Asymptotes** - **Vertical asymptotes:** where denominator is zero: $$x = -3$$ and $$x = 2$$. - **Horizontal asymptote:** Since degree of denominator (2) is greater than numerator (0), horizontal asymptote is: $$y = 0$$. 6. **Step 5: Range** The function never equals zero (horizontal asymptote), but can take all other values except possibly some values near vertical asymptotes. Because the function is continuous on intervals between vertical asymptotes and tends to $$\pm \infty$$ near them, the range is: $$(-\infty, 0) \cup (0, \infty)$$. **Final answers:** - Domain: $$(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$$ - Range: $$(-\infty, 0) \cup (0, \infty)$$ - y-intercept: $$\left(0, -\frac{1}{6}\right)$$ - No x-intercepts - Vertical asymptotes: $$x = -3$$ and $$x = 2$$ - Horizontal asymptote: $$y = 0$$