1. The problem is to graph the function $f(x) = \sqrt{x + 4}$ and plot four points on it, including the leftmost point. 2. The function $f(x) = \sqrt{x + 4}$ is defined for $x + 4 \geq 0$, so the domain is $x \geq -4$. 3. To plot points, choose values of $x$ starting from the leftmost domain point $-4$ and increasing. 4. Calculate $f(-4) = \sqrt{-4 + 4} = \sqrt{0} = 0$, so the point is $(-4, 0)$. 5. Calculate $f(-3) = \sqrt{-3 + 4} = \sqrt{1} = 1$, so the point is $(-3, 1)$. 6. Calculate $f(-2) = \sqrt{-2 + 4} = \sqrt{2} \approx 1.414$, so the point is $(-2, 1.414)$. 7. Calculate $f(-1) = \sqrt{-1 + 4} = \sqrt{3} \approx 1.732$, so the point is $(-1, 1.732)$. 8. These points can be plotted on the coordinate plane with $x$-axis from $-12$ to $12$ and $y$-axis from $-12$ to $12$. 9. The graph starts at $(-4,0)$ and increases slowly as $x$ increases. Final answer: The function $f(x) = \sqrt{x + 4}$ is graphed starting at $x = -4$ with points $(-4,0)$, $(-3,1)$, $(-2,1.414)$, and $(-1,1.732)$ plotted on the graph.