f(x) = e^(x+1). Sketch (a) y=|f(x)| (b) y=f(x−1) (c) y=2−f(x)

Since $f(x) = e^{x+1}$ is always positive ($e^{x+1} > 0$ for all $x$), the absolute value function $y = |f(x)|$ will be identical to $y = f(x)$.

*   **Equation:** $y = e^{x+1}$
*   **Key Features:**
    *   Horizontal asymptote: $y=0$
    *   Y-intercept: $(0, e)$
    *   X-intercept: None
*   **Shape:** The graph is an increasing curve that approaches $y=0$ as $x \to -\infty$ and grows without bound as $x \to \infty$. It passes through $(0, e)$.

This is a horizontal shift of $f(x)$ by 1 unit to the right.

*   **Equation:** $y = e^{(x-1)+1} = e^x$
*   **Key Features:**
    *   Horizontal asymptote: $y=0$
    *   Y-intercept: $(0, 1)$
    *   X-intercept: None
*   **Shape:** The graph is an increasing curve that approaches $y=0$ as $x \to -\infty$ and grows without bound as $x \to \infty$. It passes through $(0, 1)$.

This involves a reflection of $f(x)$ across the x-axis (due to $-f(x)$) and then a vertical shift upwards by 2 units (due to $+2$).

*   **Equation:** $y = 2 - e^{x+1}$
*   **Key Features:**
    *   Horizontal asymptote: $y=2$ (as $x \to -\infty$, $e^{x+1} \to 0$, so $y \to 2$)
    *   Y-intercept: $(0, 2-e)$
    *   X-intercept: $(\ln(2)-1, 0)$ (found by setting $2-e^{x+1}=0 \implies e^{x+1}=2 \implies x+1=\ln(2) \implies x=\ln(2)-1$)
*   **Shape:** The graph is a decreasing curve that approaches $y=2$ as $x \to -\infty$ and decreases without bound as $x \to \infty$. It passes through $(0, 2-e)$ and $(\ln(2)-1, 0)$.

The part of the graph where $f(x) \ge 0$ remains unchanged. The part where $f(x) < 0$ is reflected across the x-axis.

*   **Equation:** $y = |\ln(x+2)|$
*   **Key Features:**
    *   Vertical asymptote: $x=-2$
    *   Y-intercept: $(0, \ln(2))$
    *   X-intercept: $(-1, 0)$
*   **Shape:** For $x \ge -1$, the graph is the same as $f(x)$. For $-2 < x < -1$, the graph is reflected upwards, so it approaches $x=-2$ from the right, decreases to $(-1,0)$, and then increases for $x > -1$. The entire graph is above or on the x-axis.

This is a horizontal shift of $f(x)$ by 3 units to the right.

*   **Equation:** $y = \ln((x-3)+2) = \ln(x-1)$
*   **Key Features:**
    *   Vertical asymptote: $x=1$
    *   Y-intercept: None (domain is $x>1$)
    *   X-intercept: $(2, 0)$ (found by setting $\ln(x-1)=0 \implies x-1=1 \implies x=2$)
*   **Shape:** The graph is an increasing curve defined for $x>1$. It approaches $x=1$ from the right and grows without bound as $x \to \infty$. It passes through $(2, 0)$.

This involves a reflection of $f(x)$ across the x-axis (due to $-f(x)$) and then a vertical shift upwards by 1 unit (due to $+1$).

*   **Equation:** $y = 1 - \ln(x+2)$
*   **Key Features:**
    *   Vertical asymptote: $x=-2$
    *   Y-intercept: $(0, 1-\ln(2))$
    *   X-intercept: $(e-2, 0)$ (found by setting $1-\ln(x+2)=0 \implies \ln(x+2)=1 \implies x+2=e \implies x=e-2$)
*   **Shape:** The graph is a decreasing curve defined for $x>-2$. It approaches $x=-2$ from the right and decreases without bound as $x \to \infty$. It passes through $(0, 1-\ln(2))$ and $(e-2, 0)$.