1. **State the problem:** We need to graph the function $h(x) = (f+g)(x)$, which means for each $x$, $h(x) = f(x) + g(x)$. We are given piecewise linear graphs for $f$ and $g$ with specific points. 2. **Identify the points for $f$ and $g$:** - $f$ has points $(-2,1)$, $(0,1)$, $(1,2)$, and $(2,2)$. - $g$ has points $(-3,-2)$, $(-1,-1)$, $(1,-1)$, and $(2,0)$. 3. **Determine the domain segments for $h(x)$:** Since $f$ and $g$ are piecewise linear with given points, we use the overlapping $x$-intervals where both are defined to find $h(x)$: - Segment 1: $x \in [-2,-1]$ (between $-2$ and $-1$) - Segment 2: $x \in [-1,0]$ - Segment 3: $x \in [0,1]$ - Segment 4: $x \in [1,2]$ 4. **Find $f(x)$ and $g(x)$ on each segment by linear interpolation:** - For $x \in [-2,-1]$: - $f$ from $(-2,1)$ to $(0,1)$ is constant $1$. - $g$ from $(-3,-2)$ to $(-1,-1)$ increases linearly. - At $x=-2$, $g(-2)$ is halfway between $-2$ and $-1$, so $g(-2) = -1.5$. - At $x=-1$, $g(-1) = -1$. - So $h(-2) = f(-2)+g(-2) = 1 + (-1.5) = -0.5$. - $h(-1) = 1 + (-1) = 0$. - For $x \in [-1,0]$: - $f$ is constant $1$ from $(-2,1)$ to $(0,1)$. - $g$ from $(-1,-1)$ to $(1,-1)$ is constant $-1$. - So $h(x) = 1 + (-1) = 0$ for $x$ in $[-1,0]$. - For $x \in [0,1]$: - $f$ from $(0,1)$ to $(1,2)$ increases linearly. - $g$ from $(1,-1)$ to $(2,0)$ increases linearly. - At $x=0$, $f(0)=1$, $g(0)$ is halfway between $-1$ and $-1$ (constant), so $g(0)=-1$. - At $x=1$, $f(1)=2$, $g(1)=-1$. - So $h(0) = 1 + (-1) = 0$, $h(1) = 2 + (-1) = 1$. - For $x \in [1,2]$: - $f$ from $(1,2)$ to $(2,2)$ is constant $2$. - $g$ from $(1,-1)$ to $(2,0)$ increases linearly. - At $x=1$, $g(1)=-1$. - At $x=2$, $g(2)=0$. - So $h(1) = 2 + (-1) = 1$, $h(2) = 2 + 0 = 2$. 5. **Summarize $h(x)$ points:** - $(-2,-0.5)$ - $(-1,0)$ - $(0,0)$ - $(1,1)$ - $(2,2)$ 6. **Graph $h(x)$ using these points with linear segments connecting them.** **Final answer:** The function $h(x)$ is piecewise linear with points $(-2,-0.5)$, $(-1,0)$, $(0,0)$, $(1,1)$, and $(2,2)$ connected by straight lines.