Question: 3. Complete Function Analysis Analyze the function: $$g(x) = \frac{(x+1)^2(x-4)}{(x+1)(x^2 -16)}$$ (a) i. Factor completely and simplify if possible. ii. Find the domain. (b) i. Identify all asymptotes and holes. ii. Find all intercepts. (c) i. Determine where the function is positive/negative. ii. Sketch the graph showing all key features (including any holes).

1. **State the problem:** We analyze the function $$g(x) = \frac{(x+1)^2(x-4)}{(x+1)(x^2 -16)}$$ including factoring, domain, asymptotes, holes, intercepts, sign, and graph. 2. **Factor completely and simplify:** - Factor denominator: $$x^2 -16 = (x-4)(x+4)$$ - Rewrite $$g(x) = \frac{(x+1)^2 (x-4)}{(x+1)(x-4)(x+4)}$$ - Cancel common factors: $$g(x) = \frac{\cancel{(x+1)} (x+1) \cancel{(x-4)}}{\cancel{(x+1)} \cancel{(x-4)} (x+4)} = \frac{x+1}{x+4}$$ 3. **Find the domain:** - Original denominator zeroes: $$x+1=0 \Rightarrow x=-1$$, $$x-4=0 \Rightarrow x=4$$, $$x+4=0 \Rightarrow x=-4$$ - Domain excludes $$x=-1, 4, -4$$ 4. **Identify asymptotes and holes:** - Holes occur where factors canceled: - At $$x=-1$$ and $$x=4$$ (since these factors canceled) - Vertical asymptote where denominator zero remains: - At $$x=-4$$ - Horizontal asymptote: - Simplified function $$g(x) = \frac{x+1}{x+4}$$ has degree 1 numerator and denominator. - Horizontal asymptote is $$y=\frac{1}{1}=1$$ 5. **Find intercepts:** - x-intercept: numerator zero $$x+1=0 \Rightarrow x=-1$$ (hole, not intercept) - y-intercept: evaluate at $$x=0$$: $$g(0) = \frac{0+1}{0+4} = \frac{1}{4}$$ 6. **Determine where function is positive/negative:** - Simplified function $$g(x) = \frac{x+1}{x+4}$$ with domain excluding $$-4, -1, 4$$ - Sign depends on numerator and denominator: - Numerator zero at $$x=-1$$ (hole) - Denominator zero at $$x=-4$$ (vertical asymptote) - Test intervals: - $$(-\infty, -4)$$: numerator negative, denominator negative, fraction positive - $$(-4, -1)$$: numerator negative, denominator positive, fraction negative - $$(-1, \infty)$$: numerator positive, denominator positive, fraction positive 7. **Sketch the graph:** - Vertical asymptote at $$x=-4$$ - Holes at $$x=-1$$ and $$x=4$$ - Horizontal asymptote at $$y=1$$ - Function simplified to $$\frac{x+1}{x+4}$$ except holes - y-intercept at $$\left(0, \frac{1}{4}\right)$$ Final answer: $$g(x) = \frac{x+1}{x+4}$$ with domain $$x \neq -4, -1, 4$$, vertical asymptote at $$x=-4$$, holes at $$x=-1$$ and $$x=4$$, horizontal asymptote $$y=1$$, positive on $$(-\infty,-4) \cup (-1, \infty)$$, negative on $$(-4,-1)$$.