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.
g(x) =

Explanation:

Step1: Identify the vertex form parameters

The general form of the absolute value function is \( g(x) = a|x - h| + k \), where \((h, k)\) is the vertex of the graph. From the graph, the vertex of \( g(x) \) is at \((0, 0)\), so \( h = 0 \) and \( k = 0 \).

Step2: Determine the value of \( a \)

We can use a point on the graph to find \( a \). Let's take the point \((2, 5)\) (wait, looking at the graph, when \( x = 2 \), \( y = 5 \)? Wait, no, looking at the grid, when \( x = 2 \), the \( y \)-value is 5? Wait, no, the graph passes through \((2, 5)\)? Wait, no, let's check the graph again. Wait, the graph has a vertex at (0,0), and when \( x = 2 \), \( y = 5 \)? Wait, no, looking at the grid, each square is 1 unit. Wait, the graph at \( x = 2 \) reaches up to \( y = 5 \)? Wait, no, maybe I made a mistake. Wait, the graph: when \( x = 2 \), the \( y \)-coordinate is 5? Wait, no, let's look at the points. The graph goes through (2, 5)? Wait, no, maybe the point (2, 5) is not correct. Wait, let's take \( x = 2 \), what's \( y \)? Looking at the graph, when \( x = 2 \), the \( y \)-value is 5? Wait, no, the grid lines: the vertical axis is \( y \), horizontal is \( x \). Each small square is 1 unit. So when \( x = 2 \), the \( y \)-coordinate is 5? Wait, no, maybe the point (2, 5) is wrong. Wait, let's take \( x = 2 \), the graph is at \( y = 5 \)? Wait, no, let's check the slope. From the vertex (0,0) to (2, 5), the slope is \( \frac{5 - 0}{2 - 0} = \frac{5}{2} \)? Wait, no, maybe I misread the graph. Wait, the graph at \( x = 2 \) is at \( y = 5 \)? Wait, no, looking at the graph, when \( x = 2 \), the \( y \)-value is 5? Wait, no, the graph has a point at (2, 5)? Wait, maybe the correct point is (2, 5). Wait, but let's use the point (2, 5) to find \( a \).

Using the formula \( g(x) = a|x - h| + k \), with \( h = 0 \), \( k = 0 \), so \( g(x) = a|x| \). Plugging in \( x = 2 \), \( g(2) = 5 \):

\( 5 = a|2| \)

\( 5 = 2a \)

\( a = \frac{5}{2} \)? Wait, no, that can't be. Wait, maybe I misread the graph. Wait, looking at the graph again, when \( x = 2 \), the \( y \)-value is 5? Wait, no, the graph at \( x = 2 \) is at \( y = 5 \)? Wait, no, the grid: the vertical lines are \( y = 10, 8, 6, 4, 2, 0, -2, etc. \). Wait, when \( x = 2 \), the graph is at \( y = 5 \)? Wait, no, maybe the point is (2, 5) is incorrect. Wait, let's take \( x = 2 \), the \( y \)-coordinate is 5? Wait, no, maybe the graph is \( g(x) = 2.5|x| \), but that's a fraction. Wait, no, maybe I made a mistake. Wait, the graph: when \( x = 2 \), \( y = 5 \)? Wait, no, let's check the graph again. Wait, the graph has a vertex at (0,0), and when \( x = 2 \), \( y = 5 \)? Wait, no, the grid lines: each square is 1 unit. So from (0,0) to (2, 5), the rise is 5, run is 2, so slope is 5/2. But maybe the point is (2, 5). Wait, but let's check another point. When \( x = 4 \), \( y = 10 \). So \( g(4) = 10 \). Using \( g(x) = a|x| \), so \( 10 = a|4| \), so \( 10 = 4a \), so \( a = \frac{10}{4} = \frac{5}{2} \). Wait, but that's 2.5. But maybe the graph is \( g(x) = 2.5|x| \), but let's check the vertex form. Wait, the vertex is (0,0), so \( h = 0 \), \( k = 0 \), and \( a = \frac{5}{2} \). Wait, but maybe I misread the graph. Wait, the graph at \( x = 2 \) is at \( y = 5 \), so \( g(2) = 5 \). So \( 5 = a|2| \), so \( a = \frac{5}{2} \). Wait, but that's 2.5. But let's check the graph again. Wait, the graph: when \( x = 2 \), the \( y \)-value is 5? Wait, no, maybe the point is (2, 5) is wrong. Wait, maybe the graph is \( g(x) = 2|x| \)? No, when \( x = 2 \), \( 2|2| = 4 \), but the graph at \( x = 2 \) is at \( y = 5 \). Wait, maybe the graph is \( g(x) = \frac{5}{2}|x| \). Wait, let's confirm with \( x = 4 \): \( \frac{5}{2}|4| = \frac{5}{2} \times 4 = 10 \), which matches the graph (at \( x = 4 \), \( y = 10 \)). So that works.

Wait, but let's check the vertex form again. The general form is \( g(x) = a|x - h| + k \). We found \( h = 0 \), \( k = 0 \), and \( a = \frac{5}{2} \). Wait, but maybe I made a mistake in the point. Wait, the graph at \( x = 2 \) is at \( y = 5 \), so \( g(2) = 5 \). So \( 5 = a|2| \), so \( a = \frac{5}{2} \). So the function is \( g(x) = \frac{5}{2}|x - 0| + 0 \), which simplifies to \( g(x) = \frac{5}{2}|x| \). Wait, but let's check the graph again. Wait, the graph has a vertex at (0,0), and when \( x = 2 \), \( y = 5 \), \( x = 4 \), \( y = 10 \), so that's correct. So the value of \( a \) is \( \frac{5}{2} \), \( h = 0 \), \( k = 0 \).

Wait, but maybe I misread the graph. Wait, the graph: when \( x = 2 \), the \( y \)-value is 5? Let me check the grid again. The vertical axis: each line is 1 unit. So from 0, up to 2, 4, 6, 8, 10. So when \( x = 2 \), the graph is at \( y = 5 \), yes. So the point (2, 5) is on the graph. So using that point, we can find \( a \).

So, \( g(x) = a|x| \), plug in (2, 5):

\( 5 = a|2| \)

\( 5 = 2a \)

\( a = \frac{5}{2} \)

So the function rule is \( g(x) = \frac{5}{2}|x| \), or \( g(x) = 2.5|x| \). But let's check with \( x = 4 \): \( \frac{5}{2}|4| = 10 \), which matches the graph (at \( x = 4 \), \( y = 10 \)). So that's correct.

Wait, but maybe the graph is actually \( g(x) = 2|x| \)? No, because when \( x = 2 \), \( 2|2| = 4 \), but the graph at \( x = 2 \) is at \( y = 5 \). So that's not correct. So the correct \( a \) is \( \frac{5}{2} \).

Wait, but let's re-examine the graph. The vertex is at (0,0), and the graph passes through (2, 5) and (4, 10). So the slope from (0,0) to (2, 5) is \( \frac{5 - 0}{2 - 0} = \frac{5}{2} \), so the function is \( g(x) = \frac{5}{2}|x| \), since it's an absolute value function, the slope for \( x > 0 \) is \( \frac{5}{2} \), and for \( x < 0 \) is \( -\frac{5}{2} \), but since it's \( |x| \), it's \( \frac{5}{2}|x| \).

So putting it all together, \( h = 0 \), \( k = 0 \), and \( a = \frac{5}{2} \), so \( g(x) = \frac{5}{2}|x - 0| + 0 \), which simplifies to \( g(x) = \frac{5}{2}|x| \).

Wait, but maybe I made a mistake in the point. Let's check the graph again. The graph at \( x = 2 \): looking at the grid, the blue line at \( x = 2 \) is at \( y = 5 \)? Wait, the vertical grid lines: each line is 1 unit. So from \( y = 0 \) to \( y = 10 \), there are 10 units. So when \( x = 2 \), the \( y \)-coordinate is 5, yes. So that's correct.

So the function rule is \( g(x) = \frac{5}{2}|x| \), or in the form \( a|x - h| + k \), it's \( \frac{5}{2}|x - 0| + 0 \), so \( g(x) = \frac{5}{2}|x| \).

Wait, but let's confirm with another point. When \( x = -2 \), \( g(-2) = \frac{5}{2}|-2| = \frac{5}{2} \times 2 = 5 \), which matches the graph (at \( x = -2 \), \( y = 5 \)). So that's correct.

Answer:

\( g(x) = \frac{5}{2}|x| \) (or \( g(x) = 2.5|x| \), but as a fraction, \( \frac{5}{2}|x| \))