1. **State the problem:** Find the key features of the function $$g(x) = \frac{x^2 + x - 6}{2x^2 + 2x - 4}$$ including intercepts, asymptotes, holes, and intervals where the function is positive or negative. 2. **Factor numerator and denominator:** Numerator: $$x^2 + x - 6 = (x + 3)(x - 2)$$ Denominator: $$2x^2 + 2x - 4 = 2(x^2 + x - 2) = 2(x + 2)(x - 1)$$ 3. **Simplify the function:** $$g(x) = \frac{(x + 3)(x - 2)}{2(x + 2)(x - 1)}$$ No common factors to cancel, so no holes at this stage. 4. **Find vertical asymptotes:** Set denominator equal to zero: $$2(x + 2)(x - 1) = 0 \implies x = -2, x = 1$$ Vertical asymptotes at $$x = -2$$ and $$x = 1$$. 5. **Find horizontal asymptote:** Degree numerator = 2, degree denominator = 2. Horizontal asymptote is ratio of leading coefficients: $$y = \frac{1}{2}$$ 6. **Find y-intercept:** Evaluate $$g(0)$$: $$g(0) = \frac{0 + 0 - 6}{0 + 0 - 4} = \frac{-6}{-4} = \frac{3}{2}$$ So y-intercept is at $$(0, \frac{3}{2})$$. 7. **Find x-intercepts:** Set numerator equal to zero: $$(x + 3)(x - 2) = 0 \implies x = -3, x = 2$$ So x-intercepts at $$x = -3$$ and $$x = 2$$. 8. **Determine sign intervals:** Check signs of numerator and denominator in intervals divided by vertical asymptotes and zeros: - For $$x < -3$$: numerator positive (both factors negative, product positive), denominator positive (both factors negative, product positive), so $$g(x) > 0$$. - Between $$-3$$ and $$-2$$: numerator negative, denominator positive, so $$g(x) < 0$$. - Between $$-2$$ and $$1$$: denominator zero at $$-2$$ (vertical asymptote), so function undefined. - For $$x > 1$$: numerator positive, denominator positive, so $$g(x) > 0$$. 9. **Summary:** - Vertical asymptotes at $$x = -2$$ and $$x = 1$$. - Horizontal asymptote at $$y = \frac{1}{2}$$. - x-intercepts at $$x = -3$$ and $$x = 2$$. - y-intercept at $$(0, \frac{3}{2})$$. - Function positive on $$(-\infty, -3) \cup (1, \infty)$$. - Function negative on $$(-3, -2)$$. Note: The original message mentioned a hole at $$x = -2$$, but since denominator zero at $$x = -2$$ is not canceled by numerator, it is a vertical asymptote, not a hole. Final answer: $$g(x) = \frac{(x + 3)(x - 2)}{2(x + 2)(x - 1)}$$ Vertical asymptotes: $$x = -2, 1$$ Horizontal asymptote: $$y = \frac{1}{2}$$ Intercepts: $$(0, \frac{3}{2}), (-3, 0), (2, 0)$$ Positive intervals: $$(-\infty, -3) \cup (1, \infty)$$ Negative interval: $$(-3, -2)$$