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 vertex of \( g(x) \)

The vertex of \( f(x) = |x| \) is at \( (0,0) \). The graph of \( g(x) \) also has its vertex at \( (0,0) \), so \( h = 0 \) and \( k = 0 \) in the form \( a|x - h| + k \).

Step2: Determine the value of \( a \)

To find \( a \), we can use a point on \( g(x) \). Let's take the point \( (6, 9) \)? Wait, no, looking at the graph, when \( x = 6 \), what's \( y \)? Wait, the graph at \( x = 6 \), let's check the slope. For \( f(x) = |x| \), the slope is \( 1 \) for \( x > 0 \). Let's take a point on \( g(x) \), say \( (6, 9) \)? Wait, no, the grid: when \( x = 6 \), the \( y \)-value. Wait, the graph goes through \( (6, 9) \)? Wait, no, looking at the graph, at \( x = 6 \), the \( y \)-coordinate is 9? Wait, no, the grid lines: each square is 1 unit. Wait, the vertex is at \( (0,0) \), and when \( x = 6 \), the \( y \)-value is 9? Wait, no, maybe I miscalculated. Wait, let's take \( x = 6 \), the point is at \( (6, 9) \)? Wait, no, the graph: from \( (0,0) \) to \( (6, 9) \)? Wait, no, the line for \( x > 0 \): let's take two points. From \( (0,0) \) to \( (6, 9) \)? Wait, no, the slope: \( \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - 0}{6 - 0} = \frac{3}{2} \)? Wait, no, maybe I made a mistake. Wait, the graph: when \( x = 2 \), \( y = 3 \)? Wait, no, let's check the point \( (6, 9) \): no, the graph at \( x = 6 \), the \( y \)-coordinate is 9? Wait, the vertical axis: the top point is at \( y = 10 \), but the graph at \( x = 6 \) is at \( y = 9 \)? Wait, no, maybe the point is \( (6, 9) \). Wait, the general form is \( g(x) = a|x - h| + k \), with \( h = 0 \), \( k = 0 \), so \( g(x) = a|x| \). Let's use a point on the graph. Let's take \( x = 6 \), \( y = 9 \)? Wait, no, looking at the graph, when \( x = 6 \), the \( y \)-value is 9? Wait, the grid: each horizontal and vertical line is 1 unit. So from \( (0,0) \) to \( (6, 9) \), the slope is \( \frac{9}{6} = \frac{3}{2} \). Wait, but let's check \( x = 2 \): if \( a = \frac{3}{2} \), then \( g(2) = \frac{3}{2} \times 2 = 3 \), which matches the graph (at \( x = 2 \), \( y = 3 \)). Wait, yes, so \( a = \frac{3}{2} \)? Wait, no, wait the graph: when \( x = 6 \), \( y = 9 \), so \( 9 = a|6 - 0| + 0 \), so \( 9 = 6a \), so \( a = \frac{9}{6} = \frac{3}{2} \). Wait, but let's check another point. When \( x = -6 \), \( y = 9 \), so \( 9 = a|-6| + 0 \), so \( 6a = 9 \), \( a = \frac{3}{2} \). So the function is \( g(x) = \frac{3}{2}|x - 0| + 0 \), which simplifies to \( g(x) = \frac{3}{2}|x| \). Wait, but let's check the slope. For \( x > 0 \), the slope is \( a \), so if \( a = \frac{3}{2} \), the slope is \( \frac{3}{2} \), which matches the point \( (2, 3) \) (since \( \frac{3 - 0}{2 - 0} = \frac{3}{2} \)). Yes, that works.

Wait, but maybe I made a mistake. Let's re-examine the graph. The vertex is at \( (0,0) \), so \( h = 0 \), \( k = 0 \). Now, take a point on the graph, say \( (6, 9) \). Plugging into \( g(x) = a|x - h| + k \), we have \( 9 = a|6 - 0| + 0 \), so \( 6a = 9 \), so \( a = \frac{9}{6} = \frac{3}{2} \). So the function rule is \( g(x) = \frac{3}{2}|x - 0| + 0 \), which is \( g(x) = \frac{3}{2}|x| \).

Answer:

\( g(x) = \frac{3}{2}|x| \) (or in the form \( \frac{3}{2}|x - 0| + 0 \))