1. **Stating the problem:** We are given a piecewise function: $$h(x) = \begin{cases} 1 - x^2 & \text{if } x \in [-1, 1] \\ 0 & \text{if } x > 1 \end{cases}$$ and a new function defined as $$f(x) = \frac{1}{2} - h(x+1).$$ The problem asks us to: (a) Identify and describe the transformations from $y = h(x)$ to $y = f(x)$. (b) Graph $f(x)$ by applying transformations step-by-step. (c) Express $f(x)$ as a piecewise function like $h(x)$. --- 2. **Part (a): Identify transformations** Let's analyze $f(x) = \frac{1}{2} - h(x+1)$ starting from $y = h(x)$. Step 1: Inside the function, the input changes from $x$ to $x + 1$. This is a **horizontal shift to the left by 1 unit**. Step 2: The function is negated: it's subtracted from $\frac{1}{2}$, so $-h(x+1)$ reflects the graph **vertically across the x-axis** (because of the negative sign). Step 3: Adding $\frac{1}{2}$ outside the function shifts the whole graph **vertically upward by $\frac{1}{2}$ units**. **Order of transformations:** - First: shift left by 1 (inside the argument), - Second: reflect vertically, - Third: shift upward by $\frac{1}{2}$. --- 3. **Part (b): Sketch transformations step-by-step** We describe the effect on the graph at each step (graphs not drawn here, only explained): **Start:** Graph of $y = h(x)$ for $x \in [-1, 1]$ is a parabola with vertex at $(0, 1)$ and zeros at $x = \pm 1$. For $x > 1$, $h(x) = 0$. **Step 1: Shift left by 1 to get $y = h(x+1)$** Domain shifts from $[-1,1]$ to $[-2, 0]$ because input $x+1$ must be in $[-1,1]$ $\,\Rightarrow\, x \in [-2, 0]$. New vertex: $(x, y) = (-1, 1)$ because $h(-1+1) = h(0) = 1$; Zeroes at $x = -2$ and $x = 0$ because $h(-2+1) = h(-1) = 0$ and similarly at $x=0$. For $x > 0$, $h(x+1) = 0$. **Step 2: Reflect vertically to get $y = -h(x+1)$** The parabola opens upward (since it was downward opening originally), vertex becomes minimum at $(-1, -1)$, zeros remain at $-2$, $0$. For $x > 0$, $-h(x+1) = -0 = 0$. **Step 3: Shift up by $\frac{1}{2}$ to get $y = \frac{1}{2} - h(x+1)$** Add $\frac{1}{2}$ to all y-values: - Vertex at $(-1, -1 + \frac{1}{2}) = (-1, -\frac{1}{2})$. - Zeros at $x = -2$ and $x = 0$ shift to $\frac{1}{2}$ because originally zeros were at 0, so after reflection and shift zeros are at $y=\frac{1}{2}$. - For $x > 0$, the function is constant $f(x) = \frac{1}{2}$. Calculate actual x-intercepts of $f(x)$: To find zeros of $f(x)$, solve: $$f(x) = 0 = \frac{1}{2} - h(x+1) \implies h(x+1) = \frac{1}{2}.$$ For $x + 1 \in [-1,1]$, $$h(t) = 1 - t^2 = \frac{1}{2} \implies t^2 = \frac{1}{2} \implies t = \pm \frac{\sqrt{2}}{2}.$$ Replace $t = x + 1$: $$x + 1 = \pm \frac{\sqrt{2}}{2} \Rightarrow x = -1 \pm \frac{\sqrt{2}}{2}.$$ Check domain $x + 1 \in [-1,1]$ and these values lie in $[-1,1]$, so valid zeros are: $$x = -1 - \frac{\sqrt{2}}{2} \approx -1.707, \quad x = -1 + \frac{\sqrt{2}}{2} \approx -0.293.$$ These are the x-intercepts of $f(x)$. --- 4. **Part (c): Express $f(x)$ as a piecewise function** The domain splits according to the inside function $x+1$: - When $x + 1 \in [-1,1] \Rightarrow x \in [-2, 0]$, then $$f(x) = \frac{1}{2} - (1 - (x+1)^2) = \frac{1}{2} - 1 + (x+1)^2 = (x+1)^2 - \frac{1}{2}.$$ - When $x + 1 > 1 \Rightarrow x > 0$, then $$f(x) = \frac{1}{2} - 0 = \frac{1}{2}.$$ Hence we define: $$ f(x) = \begin{cases} (x+1)^2 - \frac{1}{2} & x \in [-2, 0] \\ \frac{1}{2} & x > 0 \end{cases} $$ **Final answer:** $$f(x) = \begin{cases} (x+1)^2 - \frac{1}{2} & x \in [-2, 0] \\ \frac{1}{2} & x > 0 \end{cases}$$