the graph of a function h is shown below. find h(1).

Explanation:

Step1: Understand the problem

We need to find the value of ( h(1) ) from the graph of the function ( h ). This means we need to find the ( y )-coordinate of the point on the graph where ( x = 1 ).

Step2: Locate ( x = 1 ) on the graph

Find the vertical line ( x = 1 ) on the coordinate plane. Then, find the point where this vertical line intersects the graph of the function ( h ).

Step3: Determine the ( y )-coordinate

From the graph, when ( x = 1 ), we look at the intersection point of ( x = 1 ) and the line of the function. By observing the grid, we can see that the line passes through the point ( (1, 2) )? Wait, no, let's check again. Wait, the line has a slope. Let's find the equation of the line. The line passes through ( (0, 1) ) (the y-intercept) and let's check another point. When ( x = -1 ), ( y = 0 ). So the slope ( m ) is ( \frac{1 - 0}{0 - (-1)} = 1 ). So the equation is ( y = x + 1 ). Then, when ( x = 1 ), ( y = 1 + 1 = 2 )? Wait, no, wait the graph: let's check the grid. Wait, the y-axis: each grid is 1 unit. When ( x = 0 ), ( y = 1 ). When ( x = 1 ), moving 1 unit to the right (since slope is 1, rise over run is 1/1), so ( y ) increases by 1. So ( y = 1 + 1 = 2 )? Wait, but let's look at the graph again. Wait, the line passes through ( (-1, 0) ), ( (0, 1) ), ( (1, 2) ), etc. So when ( x = 1 ), ( y = 2 )? Wait, no, maybe I made a mistake. Wait, the graph: let's see the coordinates. The x-axis: from -6 to 6, y-axis from -6 to 6. The line goes through (0,1), and when x=1, the point is at (1,2)? Wait, but let's check the grid. Each square is 1 unit. So at x=1, the y-coordinate is 2? Wait, no, maybe I misread. Wait, the line: when x=0, y=1. Then, for each x increase by 1, y increases by 1. So x=1, y=2. So ( h(1) = 2 ).

Wait, but let's confirm. Let's plot the points. When x=1, the vertical line x=1 intersects the graph at (1, 2). So the y-coordinate is 2. So ( h(1) = 2 ).

Answer:

( \boxed{2} )