1. **Problem Statement:** Given the function $f(x) = \frac{3}{x+1}$, find: i) The y-intercept. ii) The vertical and horizontal asymptotes. iii) The intervals where $f(x)$ is increasing or decreasing. iv) The intervals where $f(x)$ is concave up or concave down. v) Sketch the graph of $f(x)$. 2. **Y-intercept:** The y-intercept occurs where $x=0$. Substitute $x=0$ into $f(x)$: $$f(0) = \frac{3}{0+1} = 3$$ So, the y-intercept is $(0, 3)$. 3. **Asymptotes:** - **Vertical asymptote:** Occurs where the denominator is zero. Set $x+1=0 \Rightarrow x=-1$. So, vertical asymptote at $x=-1$. - **Horizontal asymptote:** For rational functions where degree of numerator is less than denominator, horizontal asymptote is $y=0$. So, horizontal asymptote at $y=0$. 4. **Increasing/Decreasing intervals:** Find the first derivative: $$f'(x) = \frac{d}{dx} \left( \frac{3}{x+1} \right) = -\frac{3}{(x+1)^2}$$ Since $(x+1)^2 > 0$ for all $x \neq -1$, and numerator is negative, $f'(x) < 0$ everywhere except at $x=-1$ where undefined. Therefore, $f(x)$ is decreasing on $(-\infty, -1)$ and $(-1, \infty)$. 5. **Concavity:** Find the second derivative: $$f''(x) = \frac{d}{dx} \left(-\frac{3}{(x+1)^2} \right) = -3 \cdot \frac{d}{dx} \left( (x+1)^{-2} \right) = -3 \cdot (-2)(x+1)^{-3} = \frac{6}{(x+1)^3}$$ - For $x < -1$, $(x+1)^3 < 0$, so $f''(x) < 0$ (concave down). - For $x > -1$, $(x+1)^3 > 0$, so $f''(x) > 0$ (concave up). 6. **Graph sketch description:** The graph has a vertical asymptote at $x=-1$ and a horizontal asymptote at $y=0$. The function decreases on both sides of $x=-1$. It is concave down on $(-\infty, -1)$ and concave up on $(-1, \infty)$. **Final answers:** - i) Y-intercept: $(0, 3)$ - ii) Vertical asymptote: $x=-1$, Horizontal asymptote: $y=0$ - iii) Decreasing on $(-\infty, -1)$ and $(-1, \infty)$ - iv) Concave down on $(-\infty, -1)$, concave up on $(-1, \infty)$ - v) Graph as described above.