Question

the graph shows g(x), which is a transformation of f(x) = |x|. write the function rule for g(x). write your answer in the form a|x - h| + k, where a, h, and k are integers or simplified fractions.

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, the vertex of \(g(x)\) is at \((1,0)\) (wait, no, looking at the graph again, the vertex is at \((1,0)\)? Wait, no, let's check the graph. The graph of \(g(x)\) has its vertex at \((1,0)\)? Wait, no, when \(x = 1\), \(y=0\)? Wait, no, looking at the grid, the vertex is at \((1,0)\)? Wait, no, let's see the graph: the point where the two lines meet is at \(x = 1\), \(y = 0\)? Wait, no, the graph crosses the x - axis at \(x = 1\)? Wait, no, looking at the graph, the vertex (the minimum point) is at \((1,0)\)? Wait, no, let's check the coordinates. Let's take two points on the right - hand side of the vertex. For example, when \(x = 3\), \(y=2\); when \(x = 5\), \(y = 4\); when \(x=9\), \(y = 8\). The slope of the right - hand line: slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2-0}{3 - 1}=\frac{2}{2}=1\)? Wait, no, when \(x = 1\), \(y = 0\); when \(x=2\), \(y = 1\); when \(x = 4\), \(y=3\)? Wait, no, the graph as shown: let's re - examine. Wait, the standard form of the absolute - value function is \(g(x)=a|x - h|+k\), where \((h,k)\) is the vertex. From the graph, the vertex is at \((1,0)\)? Wait, no, looking at the graph, the vertex is at \((1,0)\)? Wait, no, the graph's vertex is at \((1,0)\)? Wait, no, let's check the y - intercept. When \(x = 0\), \(y = 1\)? Wait, no, the graph at \(x = 0\) has \(y = 1\)? Wait, no, the original function is \(f(x)=|x|\), which has vertex at \((0,0)\). The transformed function \(g(x)\): let's find the vertex. Looking at the graph, the vertex is at \((1,0)\)? Wait, no, the graph shows that the vertex is at \((1,0)\)? Wait, no, let's take the left - hand side. When \(x=-1\), \(y = 2\); when \(x=-3\), \(y = 4\); when \(x=-9\), \(y = 10\). The slope of the left - hand line: \(\frac{y_2 - y_1}{x_2 - x_1}=\frac{2-0}{-1 - 1}=\frac{2}{-2}=-1\). Wait, the vertex is at \((1,0)\)? Wait, no, maybe I made a mistake. Wait, let's look at the graph again. The vertex (the point where the two linear pieces meet) is at \((1,0)\). Now, the general form is \(g(x)=a|x - h|+k\), with \((h,k)=(1,0)\). Now, let's find \(a\). Let's take a point on the graph, say \((3,2)\). Substitute into \(g(x)=a|x - 1|+0\). So \(2=a|3 - 1|=a\times2\), so \(a = 1\). Wait, but let's check another point. When \(x = 9\), \(g(9)=a|9 - 1|=8a\). From the graph, when \(x = 9\), \(y = 8\), so \(8a=8\), so \(a = 1\). Wait, but let's check the left - hand side. When \(x=-1\), \(g(-1)=a|-1 - 1|=2a\). From the graph, when \(x=-1\), \(y = 2\), so \(2a = 2\), so \(a = 1\). Wait, but the original function \(f(x)=|x|\) has vertex at \((0,0)\) and \(a = 1\). But this graph has vertex at \((1,0)\) and \(a = 1\). Wait, no, maybe the vertex is at \((1,0)\)? Wait, no, let's check the graph again. Wait, the graph: when \(x = 0\), \(y = 1\); when \(x = 1\), \(y = 0\); when \(x = 2\), \(y = 1\); when \(x=3\), \(y = 2\), etc. So the function is \(g(x)=|x - 1|+0\)? Wait, no, when \(x = 0\), \(g(0)=|0 - 1|=1\), which matches the graph (at \(x = 0\), \(y = 1\)). When \(x=-1\), \(g(-1)=|-1 - 1|=2\), which matches the graph (at \(x=-1\), \(y = 2\)). When \(x = 9\), \(g(9)=|9 - 1|=8\), which matches the graph (at \(x = 9\), \(y = 8\)). So the vertex is at \((h,k)=(1,0)\), \(a = 1\), \(k = 0\). Wait, but let's confirm the form. The general form is \(g(x)=a|x - h|+k\). So \(h = 1\), \(k = 0\), \(a = 1\). So \(g(x)=|x - 1|+0\), which simplifies to \(g(x)=|x - 1|\). Wait, but let's check the slope. The right - hand side: for \(x\geq1\), \(g(x)=x - 1\) (since \(a = 1\), \(x-1\) when \(x\geq1\)), slope is 1. The left - hand side: for \(x\lt1\), \(g(x)=-(x - 1)=-x + 1\), slope is - 1. Which matches the graph's slopes (left - hand slope is - 1, right - hand slope is 1). So the function rule is \(g(x)=|x - 1|\), or in the form \(a|x - h|+k\), \(a = 1\), \(h = 1\), \(k = 0\), so \(g(x)=1|x - 1|+0\).

Wait, maybe I made a mistake in the vertex. Let's re - examine the graph. The vertex is at \((1,0)\)? Wait, the graph shows that the minimum point is at \(x = 1\), \(y = 0\). Let's check the point \(x = 1\), \(y = 0\): yes, that's the vertex. Then, using the point - slope form. For the right - hand line (where \(x\geq1\)), the equation is \(y-0=m(x - 1)\). We can take a point on the right - hand line, say \((3,2)\). Then \(m=\frac{2 - 0}{3 - 1}=1\), so the equation is \(y=x - 1\) for \(x\geq1\). For the left - hand line (where \(x\lt1\)), take a point \((-1,2)\). The slope \(m=\frac{2-0}{-1 - 1}=\frac{2}{-2}=-1\), so the equation is \(y-0=-1(x - 1)\), which is \(y=-x + 1\) for \(x\lt1\). Combining these two, we get \(g(x)=|x - 1|\), which is \(g(x)=1|x - 1|+0\).

Step1: Find the vertex \((h,k)\)

The vertex of the absolute - value function \(g(x)=a|x - h|+k\) is the point where the two linear parts meet. From the graph, the vertex is at \((h,k)=(1,0)\).

Step2: Find the value of \(a\)

We can use a point on the graph to find \(a\). Let's use the point \((3,2)\) (on the right - hand side of the vertex). Substitute \(x = 3\), \(y = 2\), \(h = 1\), and \(k = 0\) into the equation \(y=a|x - h|+k\):
\[2=a|3 - 1|+0\]
\[2 = 2a\]
Divide both sides by 2: \(a = 1\).

Step3: Write the function rule

Substitute \(a = 1\), \(h = 1\), and \(k = 0\) into the form \(g(x)=a|x - h|+k\):
\[g(x)=1|x - 1|+0\]
Simplifying, we get \(g(x)=|x - 1|\) or \(g(x)=1|x - 1|+0\).

Answer:

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