The equation provided is:

$y = 2|x + 3|$

This is an absolute value function, which creates a "V"-shaped graph. Let's go step by step to graph it:

Step-by-Step Graphing Process:

1. Basic Form: The given equation is $y = 2 |x + 3|$, which is in the form $y = a|x - h| + k$, where:

   • $a = 2$ (this controls the slope or steepness of the "V" shape),
   • $h = -3$ (this is a horizontal shift; the graph is shifted 3 units to the left),
   • $k = 0$ (there is no vertical shift in this case).

2. Vertex: The vertex (the point where the graph "V" bends) occurs at $x = -3$. So, the vertex is at $(-3, 0)$.

3. Shape and Direction: The absolute value creates two linear parts:

   • For $x \geq -3$, the equation behaves like $y = 2(x + 3)$,
   • For $x < -3$, the equation behaves like $y = 2(-(x + 3)) = -2(x + 3)$.

4. Slope: The slope is 2, meaning that for every 1 unit you move horizontally, the graph moves 2 units vertically.

Key Points:

• At $x = -3$: The vertex is at $y = 0$.
• At $x = -2$: Plugging into the equation gives $y = 2|-2 + 3| = 2 \times 1 = 2$.
• At $x = 0$: Plugging into the equation gives $y = 2|0 + 3| = 2 \times 3 = 6$.
• At $x = -4$: Plugging into the equation gives $y = 2|-4 + 3| = 2 \times 1 = 2$.
• At $x = -6$: Plugging into the equation gives $y = 2|-6 + 3| = 2 \times 3 = 6$.

Plot:

1. Plot the vertex at $(-3, 0)$.
2. On the right side (for $x \geq -3$), plot points like $(-2, 2)$, $(0, 6)$, etc., with a slope of 2.
3. On the left side (for $x < -3$), plot points like $(-4, 2)$, $(-6, 6)$, etc., with a slope of -2.

The graph will form a "V" shape, with the vertex at $(-3, 0)$, and the lines extending upwards on both sides.

Would you like any further clarifications or help with this? Here are some follow-up questions:

1. How does the slope of 2 affect the steepness of the graph compared to a slope of 1?
2. What happens if we change the value of $h$ in the equation?
3. How would the graph change if the absolute value term had a negative sign in front of it?
4. Can you explain why the graph forms a "V" shape instead of a smooth curve?
5. What would the graph look like if we added a constant $k$ to the equation?

Tip: Always plot the vertex first when dealing with absolute value functions, as it helps anchor the graph.