Question

evaluate the function graphically.
find ( f(3) )

Explanation:

Step1: Locate x = 3 on x - axis

Find the point where \( x = 3 \) on the horizontal (x - axis) of the graph.

Step2: Find the corresponding y - value on the graph

From \( x = 3 \), move vertically to intersect the graph of the function. The open circle at \( x = 3 \) is part of the function's graph (the solid dot below is a different point, not part of the function's graph here). The y - value at the open circle (which is on the line) can be determined. Looking at the line, when \( x = 3 \), the y - value is 2? Wait, no, let's check the line. The line crosses the y - axis at 5 (when \( x = 0 \), \( y = 5 \)) and crosses the x - axis at 5 (when \( y = 0 \), \( x = 5 \)). So the equation of the line is \( y=-x + 5 \) (since slope \( m=\frac{0 - 5}{5 - 0}=-1 \), and y - intercept \( b = 5 \)). Plugging \( x = 3 \) into \( y=-x + 5 \), we get \( y=-3 + 5=2 \). Wait, but the open circle is at \( x = 3 \), what's the y - value? Wait, maybe I misread. Wait the graph: the line goes from top left, crosses y - axis at 5, then at \( x = 3 \), there's an open circle. Let's see the coordinates. The open circle is at (3, 2)? Wait no, let's check the grid. The x - axis: from 0 to 10, y - axis from - 10 to 10. Wait the line: when x = 0, y = 5; when x = 5, y = 0. So the slope is - 1. So at x = 3, y=5 - 3=2. So the open circle is at (3, 2)? Wait but the solid dot is below, but the function's graph is the line with the open circle (since the solid dot is not on the line). So to find f(3), we look at the function's graph (the line), at x = 3, the y - value is 2 (from the open circle, which is part of the function's graph). Wait, maybe I made a mistake. Wait the problem says "evaluate the function graphically". So we look at the function's graph (the line with the open circle, not the solid dot). So at x = 3, the function's graph (the line) has a point (open circle) at y = 2? Wait no, maybe the open circle is at (3, 2)? Wait let's check the coordinates again. The x - axis: each grid is 1 unit. So x = 3, moving up to the line: the line at x = 3, what's the y? Let's see, the line passes through (0,5) and (5,0). So the equation is y=-x + 5. So when x = 3, y = 2. So f(3)=2? Wait but the open circle is at (3, 2)? Wait maybe the open circle is at (3, 2), so f(3) is 2? Wait no, maybe I messed up. Wait the graph: the open circle is at x = 3, what's the y - coordinate? Let's count the grid. From x = 0, y = 5 (top of the line at y - axis). Then at x = 1, y = 4; x = 2, y = 3; x = 3, y = 2. Yes, so the open circle is at (3, 2), so f(3)=2. Wait but the solid dot is at (3, - 3) or something? No, the solid dot is below, but that's not part of the function. The function's graph is the line with the open circle (since the line is the function, and the open circle is a point on the function, maybe a removable discontinuity, but for evaluating f(3), we use the function's graph, which is the line. So f(3) is the y - value when x = 3 on the function's graph (the line), which is 2. Wait, no, wait the open circle: is the open circle at (3, 2) or (3, 1)? Wait looking at the graph, the open circle is at x = 3, and the y - value is 2? Wait maybe I made a mistake. Wait let's re - examine. The line: when x = 3, the y - coordinate on the line is 2. So f(3)=2.

Wait, maybe the initial thought was wrong. Let's do it again. The line has a y - intercept at 5 (x = 0, y = 5) and x - intercept at 5 (y = 0, x = 5). So slope m=(0 - 5)/(5 - 0)=-1. Equation: y=-x + 5. So when x = 3, y=-3 + 5=2. So f(3)=2. The open circle is at (3, 2), so that's the value of the function at x = 3. The solid dot below is a different point, not part of the function's graph for f(3).

Answer:

2