Q. Which graph represents the function \(f(x) = |x + 3|\)?

Solution — step-by-step

1. Recall the definition of absolute value.

   For any expression u,
   \[ |u| = \begin{cases}
   u & \text{if } u \ge 0,\\[4pt]
   -u & \text{if } u < 0. \end{cases} \]

   We will apply this with u = x + 3.

2. Write f as a piecewise linear function.

   Substitute u = x + 3 into the definition:
   \[ f(x)=|x+3| = \begin{cases}
   x+3 & \text{if } x+3 \ge 0 \;\text{(i.e. } x \ge -3\text{)},\\[4pt]
   -(x+3) & \text{if } x+3 < 0 \;\text{(i.e. } x < -3\text{)}. \end{cases} \] Simplifying the second branch gives \[ f(x)=\begin{cases} x+3 & \text{if } x \ge -3,\\[4pt] -x-3 & \text{if } x < -3. \end{cases} \]

   So the graph is made of two straight-line pieces meeting at x = -3.

3. Find the vertex (the corner point).

   The corner occurs where the inside of the absolute value is zero: x+3 = 0, so x = -3. Evaluate f there:
   \[ f(-3) = |{-3}+3| = 0. \]
   Thus the vertex is at the point (-3, 0).

4. Determine slopes and directions of the two arms.

   For x \ge -3 the formula is f(x) = x + 3, which is a line of slope 1. For x < -3 the formula is f(x) = -x - 3, which is a line of slope -1. Therefore the graph is a V-shape with the right arm rising with slope 1 and the left arm rising to the left with slope -1.

5. Check a few points to fix placement.

   Evaluate f at some sample x-values:
   \[ f(0)=|0+3|=3 \quad\text{so }(0,3) \text{ is on the graph}, \]
   \[ f(-2)=|-2+3|=1 \quad\text{so }(-2,1) \text{ is on the graph}, \]
   \[ f(-4)=|-4+3|=1 \quad\text{so }(-4,1) \text{ is on the graph}. \]
   These are consistent with a V whose vertex is at (-3,0) and symmetric about the vertical line x = -3.

6. State domain and range.

   Domain: all real numbers, written \(\mathbb{R}\).
   Range: all real y with y \ge 0, written [0, \infty).

7. Conclusion — which graph represents f(x)=|x+3|?

   The correct graph is the V-shaped graph with vertex at (-3,0), symmetric about the vertical line x = -3, with the right arm the line y = x + 3 (slope 1) for x \ge -3 and the left arm the line y = -x – 3 (slope -1) for x < -3. It passes through points such as (-4,1), (-2,1), and (0,3), and its y-values are all nonnegative.