Question

the graph of a function f is shown below. find f(3).

Explanation:

Step1: Recall function graph interpretation

To find \( f(3) \), we need to determine the \( y \)-value of the function \( f \) when \( x = 3 \). This means we look at the point on the graph of \( f \) where the \( x \)-coordinate is 3 and find the corresponding \( y \)-coordinate.

Step2: Analyze the graph

Looking at the given graph, we identify the line representing the function \( f \). We then find the point on this line where \( x = 3 \). By examining the grid and the slope of the line (or directly reading the coordinates), we can see that when \( x = 3 \), the \( y \)-coordinate (the value of \( f(3) \)) is -3. Wait, let's re - check. Wait, the line passes through (0,2) and (1,0)? Wait, no, let's find the equation of the line. The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points: when \( x = 0 \), \( y = 2 \); when \( x = 1 \), \( y = 0 \). So slope \( m=\frac{0 - 2}{1 - 0}=- 2\). The equation of the line is \( y=mx + b\), with \( b = 2 \) (y - intercept). So \( y=-2x + 2\). Now, when \( x = 3 \), \( y=-2(3)+2=-6 + 2=-4\)? Wait, maybe my initial points are wrong. Let's look at the graph again. The line crosses the x - axis at \( x = 1 \) (since when \( y = 0 \), \( x = 1 \)) and the y - axis at \( y = 2 \) (when \( x = 0 \)). Wait, no, when \( x=3\), let's count the grid. Each square is 1 unit. From \( x = 0 \), moving 3 units to the right (x = 3), and since the slope is - 2 (for each 1 unit right, 2 units down). From \( x = 0,y = 2 \), at \( x = 1,y = 0 \) (down 2), \( x = 2,y=-2 \) (down 2), \( x = 3,y=-4 \)? Wait, maybe I made a mistake earlier. Wait, let's check the graph again. The line: when x = 3, what is the y - value? Let's see the graph. The line goes from the top left to bottom right. At x = 0, y = 2; x = 1, y = 0; x = 2, y=-2; x = 3, y=-4? Wait, no, maybe the slope is - 1? Wait, no, let's re - examine the graph. Wait, the y - axis: from 0, up is positive, down is negative. The x - axis: right is positive, left is negative. Let's take two points: (0,2) and (2, - 2). Then slope \( m=\frac{-2 - 2}{2 - 0}=\frac{-4}{2}=-2\). So equation \( y=-2x + 2\). So when \( x = 3 \), \( y=-2(3)+2=-4 \). Wait, but maybe the graph is such that at x = 3, the y - value is - 4? Wait, maybe my first analysis was wrong. Let's look at the graph again. The line: when x = 3, looking at the vertical line x = 3, where does it intersect the function's graph? The graph is a straight line. Let's count the units. From x = 0 (y = 2) to x = 3: 3 units to the right. Since the slope is - 2 (each x unit right, y decreases by 2), so 3 units right, y decreases by 6? No, 2 per unit. So 3*2 = 6, 2-6=-4. So \( f(3)=-4 \).

Wait, maybe I messed up the slope. Let's take another approach. Let's look at the graph: when x = 3, the point on the line has a y - coordinate of - 4? Wait, no, maybe the correct answer is - 4? Wait, let's check with the graph. If we draw a vertical line at x = 3, it intersects the line at (3, - 4). So \( f(3)=-4 \).

Answer:

\( f(3)=\boxed{-4} \)