Question

evaluate the function graphically.
find $f(-5)$

Explanation:

Step1: Locate x = -5 on x - axis

Find the vertical line corresponding to \( x=-5 \) on the coordinate plane.

Step2: Determine the function's value at x = -5

The line segment (the straight - line part of the graph) passes through \( x = - 5 \). We can also use the equation of the straight - line part of the graph. The y - intercept of the straight line is 2 (when \( x = 0 \), \( y=2 \)) and the slope \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points on the straight line, for example, \( (0,2) \) and \( (2,0) \). The slope \( m=\frac{0 - 2}{2-0}=- 1 \). So the equation of the line is \( y=-x + 2 \). When \( x=-5 \), substitute \( x=-5 \) into the equation: \( y=-(-5)+2=5 + 2=7 \)? Wait, no, wait. Wait, looking at the graph, the straight line: when \( x=-5 \), let's check the graph again. Wait, the straight line has a y - intercept at (0,2) and goes down with a slope of - 1? Wait, no, when \( x = 2 \), \( y=0 \), when \( x=-5 \), let's calculate \( y\): using \( y=mx + b \), \( m=\frac{0 - 2}{2-0}=-1 \), \( b = 2 \), so \( y=-x + 2 \). When \( x=-5 \), \( y=-(-5)+2=5 + 2 = 7 \)? But wait, there are open circles, but at \( x=-5 \), which part of the graph is defined? The straight - line part (the line with the arrow going down) is defined at \( x=-5 \) (since the parabola's domain and the straight line's domain: the parabola is on the left, but at \( x=-5 \), the straight line is the one that is defined there? Wait, no, let's look at the graph again. Wait, the straight line passes through \( x=-5 \). Wait, maybe I made a mistake in the slope. Wait, when \( x = 0 \), \( y = 2 \); when \( x=1 \), \( y = 1 \); when \( x=-5 \), \( y=-(-5)+2=7 \)? But wait, the open circles are at some other x - values. Wait, the problem is to find \( f(-5) \) graphically. So we find \( x=-5 \) on the x - axis, then move up or down to the graph. The straight - line part of the graph (the line with the arrow going down) passes through \( x=-5 \). Let's check the y - value: when \( x=-5 \), the y - value from the straight line (using the line equation or just reading from the graph) is 7? Wait, no, wait, maybe the line equation is wrong. Wait, let's take two points on the straight line: (0,2) and (2,0). The slope is \( (0 - 2)/(2 - 0)=-1 \), so equation is \( y=-x + 2 \). When \( x=-5 \), \( y=-(-5)+2=7 \). But wait, is there a point at \( x=-5 \) on the straight line? Yes, because the straight line is a continuous part (except for the open circles at some x - values, but \( x=-5 \) is not one of the open - circle x - values). So \( f(-5)=7 \)? Wait, no, wait, maybe I messed up. Wait, the graph: the straight line, when \( x=-5 \), let's count the grid. Each grid is 1 unit. From \( x = 0 \), moving left 5 units to \( x=-5 \), moving up 5 units from \( y = 2 \) (since slope is - 1, so for each unit left, y increases by 1). So from \( x = 0 \), \( y = 2 \), moving left 5 units (to \( x=-5 \)), y increases by 5, so \( y=2 + 5=7 \). So \( f(-5)=7 \).

Wait, no, wait, maybe the line is \( y=-x + 2 \), so when \( x=-5 \), \( y = 5 + 2=7 \). So the value of \( f(-5) \) is 7.

Answer:

\( f(-5)=7 \)