Question

which of the following is the graph of this absolute value function?
y = |x + 1|

Explanation:

Step1: Recall the vertex form of absolute value functions

The general form of an absolute value function is \( y = |x - h| + k \), where \((h, k)\) is the vertex of the V - shaped graph. For the function \( y=|x + 1|\), we can rewrite it as \( y=|x-(- 1)|+0 \). So, the vertex of the graph should be at \((h,k)=(-1,0)\).

Step2: Analyze the graphs

• For the first graph: Check the vertex. The vertex appears to be at \((-1,0)\). Let's verify with a point. When \(x = 0\), \(y=|0 + 1|=1\)? Wait, no, when \(x = 0\), \(y=|0 + 1| = 1\)? Wait, no, the first graph: when \(x=0\), the \(y\) - value is 1? Wait, no, let's re - check. The first graph has a vertex at \((-1,0)\). When \(x=-1\), \(y = |-1 + 1|=0\) (correct for vertex). When \(x = 0\), \(y=|0 + 1| = 1\), and when \(x=-2\), \(y=|-2 + 1|=1\). The slope of the right - hand side (where \(x\geq - 1\)): the function \(y=x + 1\) (for \(x\geq - 1\)) has a slope of 1, and the left - hand side (where \(x\lt - 1\)): the function \(y=-(x + 1)=-x - 1\) has a slope of - 1.
• The second graph: The vertex is at \((0,1)\), which does not match the vertex \((-1,0)\) of \(y = |x + 1|\).
• The third graph: The vertex is at \((1,0)\), which does not match the vertex \((-1,0)\) of \(y = |x + 1|\).

Answer:

The first graph (the one with vertex at \((-1,0)\), passing through \((0,1)\) and \((-2,1)\) with a V - shape opening upwards)