Question

write your answer in the form a|x - h| + k, where a, h, and k are integers or simplified fractions.
a(x) =

Explanation:

Step1: Identify the vertex

The vertex of the absolute - value function \(y = a|x - h|+k\) is at the point \((h,k)\). From the graph, we can see that the vertex is at \((0,1)\)? Wait, no, looking at the graph again, the vertex is at \((0,1)\)? Wait, no, the graph has its minimum point at \((0,1)\)? Wait, no, when \(x = 0\), \(y=1\)? Wait, no, looking at the grid, when \(x = 0\), the \(y\) - coordinate is \(1\)? Wait, no, the graph shows that at \(x = 0\), the \(y\) - value is \(1\)? Wait, no, let's check the grid again. The vertical axis (y - axis) has marks at 0, 2, 4, etc. Wait, the vertex is at \((0,1)\)? Wait, no, the graph intersects the y - axis at \((0,1)\)? Wait, no, looking at the graph, when \(x = 0\), the \(y\) - value is \(1\)? Wait, no, the graph is a V - shaped graph with the vertex at \((0,1)\)? Wait, no, let's look at the points. For \(x = 2\), \(y = 3\); for \(x=-2\), \(y = 3\). Wait, the vertex form of an absolute - value function is \(y=a|x - h|+k\), where \((h,k)\) is the vertex. From the graph, the vertex is at \((h,k)=(0,1)\)? Wait, no, when \(x = 0\), \(y = 1\)? Wait, no, the graph shows that at \(x = 0\), the \(y\) - coordinate is \(1\)? Wait, no, let's calculate the slope. For the right - hand side of the V (where \(x\geq0\)), we can take two points. Let's take \((0,1)\) and \((2,3)\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3 - 1}{2 - 0}=\frac{2}{2}=1\). For the left - hand side (where \(x<0\)), take \((-2,3)\) and \((0,1)\), the slope \(m=\frac{1 - 3}{0-(-2)}=\frac{-2}{2}=- 1\). The general form of the absolute - value function is \(y=a|x - h|+k\). Since the vertex is at \((h,k)=(0,1)\) and the slope of the right - hand side is \(a = 1\) (because for \(x\geq h\), \(y=a(x - h)+k\), and the slope of this linear part is \(a\)).

Wait, let's re - examine the graph. Wait, the user's graph: when \(x = 0\), \(y = 1\)? Wait, no, looking at the original graph, the vertex is at \((0,1)\)? Wait, no, the graph has a minimum at \((0,1)\). Let's take two points on the right side: \((0,1)\) and \((2,3)\). The change in \(y\) is \(3 - 1=2\), the change in \(x\) is \(2-0 = 2\), so the slope \(a=\frac{2}{2}=1\). So the equation of the right - hand line (for \(x\geq0\)) is \(y-1 = 1\times(x - 0)\), which is \(y=x + 1\). For the left - hand line (for \(x<0\)), take \((-2,3)\) and \((0,1)\). The slope is \(\frac{1 - 3}{0-(-2)}=\frac{-2}{2}=-1\), and the equation is \(y - 1=-1\times(x - 0)\), which is \(y=-x + 1\). We can write this as \(y = |x|+1\)? Wait, no, when \(x = 2\), \(|2|+1=3\), which matches. When \(x=-2\), \(|-2|+1 = 3\), which matches. When \(x = 0\), \(|0|+1=1\), which matches. Wait, but in the graph, when \(x = 0\), the \(y\) - value is 1? Wait, the original graph: the vertical axis has marks at 0, 2, 4, etc. Wait, maybe I made a mistake. Wait, let's look at the graph again. The vertex is at \((0,1)\)? Wait, no, the graph shows that at \(x = 0\), the \(y\) - coordinate is 1. Let's check the point \((2,3)\): \(|2|+1 = 3\), correct. \((-2,3)\): \(|-2|+1=3\), correct. So the function is \(y = |x|+1\), which is in the form \(a|x - h|+k\) with \(a = 1\), \(h = 0\), and \(k = 1\). Wait, but let's check another point. When \(x = 4\), \(y=|4| + 1=5\), which should be on the graph. Looking at the graph, when \(x = 4\), the \(y\) - value is 5, which matches. When \(x=-4\), \(y = |-4|+1 = 5\), which also matches. So the function is \(a(x)=|x|+1\), or in the form \(a|x - h|+k\), it is \(1|x - 0|+1\).

Step2: Confirm the values of \(a\), \(h\), and \(k\)

- The vertex of the absolute - value function \(y=a|x - h|+k\) is \((h,k)\). From the graph, the vertex is at \((0,1)\), so \(h = 0\) and \(k = 1\).
- To find \(a\), we can use a point on the graph. Let's use the point \((2,3)\). Substitute \(x = 2\), \(y = 3\), \(h = 0\), and \(k = 1\) into the equation \(y=a|x - h|+k\):
\(3=a|2 - 0|+1\)
\(3 = 2a+1\)
Subtract 1 from both sides: \(2a=3 - 1=2\)
Divide both sides by 2: \(a = 1\)

Answer:

\(a(x)=1|x - 0|+1\) or \(a(x)=|x| + 1\)