evaluating functions, a graph - notes
use the graph for 14 - 17.
14. find f (0) =
15. find f (-4) =
16. find f (2) =
17. if f (x) = 5. what is x?
x =
use the graph below for problems 18 - 21
18. find g (3) =
19. find f (4) =
20. find f (g (3))

Question

Accepted Answer

### Question 14: Find $ f(0) $
# Explanation:
## Step 1: Locate $ x = 0 $ on the graph
Find the point on the graph where the $ x $-coordinate is 0. From the graph, at $ x = 0 $, the $ y $-coordinate (which is $ f(0) $) is 0.
# Answer:
$ f(0) = 0 $

### Question 15: Find $ f(-4) $
# Explanation:
## Step 1: Locate $ x = -4 $ on the graph
Find the point on the graph where the $ x $-coordinate is -4. The line on the left side passes through $ x = -4 $. To find the $ y $-value, we can see the slope or use the shape. The left line goes from $ (-5, 5) $ to $ (0, 0) $, so the equation of the left line (for $ x \leq 0 $) can be found, but visually, at $ x = -4 $, the $ y $-coordinate is 4 (since from $ x = -5 $ (y=5) to $ x = 0 $ (y=0), the change in y is -5 for change in x of 5, so slope is -1. So $ y - 5 = -1(x + 5) $, so $ y = -x $. So at $ x = -4 $, $ y = -(-4) = 4 $.
# Answer:
$ f(-4) = 4 $

### Question 16: Find $ f(2) $
# Explanation:
## Step 1: Locate $ x = 2 $ on the graph
Find the point on the graph where the $ x $-coordinate is 2. There is a closed dot at $ x = 2 $ with $ y $-coordinate 2.
# Answer:
$ f(2) = 2 $

### Question 17: If $ f(x) = 5 $, what is $ x $?
# Explanation:
## Step 1: Locate $ y = 5 $ on the graph
Find the points where the $ y $-coordinate is 5. The left line (for $ x \leq 0 $) has the equation $ y = -x $ (from earlier). Setting $ y = 5 $, we get $ 5 = -x $, so $ x = -5 $. Let's check the graph: at $ x = -5 $, the left line has a point (the arrow starts there), so $ y = 5 $ when $ x = -5 $.
# Answer:
$ x = -5 $

### Question 18: Find $ G(3) $
# Explanation:
## Step 1: Locate $ x = 3 $ on the $ G $ graph (the light - colored graph)
Looking at the graph for $ G $ (the non - dark line), at $ x = 3 $, we can see the point. From the graph, the $ G $ graph at $ x = 3 $ has a $ y $-coordinate. Wait, the $ G $ graph: let's see the points. The $ G $ graph has a point around $ x = 2 $ (open circle, but the closed dot is at $ y = 6 $? Wait, maybe better to look at the coordinates. Wait, the $ G $ graph: at $ x = 3 $, looking at the line, the $ y $-value. Wait, the $ G $ graph (light) has a point at $ x = 3 $? Wait, maybe the $ G $ graph: let's see the first part: from $ x = 2 $ (open circle at $ y = 6 $, closed dot at $ y = 6 $? Wait, no, the $ G $ graph: the left part has a point at $ x = 2 $ (open circle, $ y = 6 $) and a point at $ x = 3 $? Wait, maybe I misread. Wait, the $ G $ graph (light) has a point at $ x = 3 $? Wait, actually, looking at the $ G $ graph (the one with the open circle at $ x = 2 $, $ y = 6 $ and a closed dot at $ x = 2 $, $ y = 6 $? No, maybe the $ G $ graph: the line from $ x = 2 $ (open circle, $ y = 6 $) goes down? Wait, no, the $ G $ graph: let's see the coordinates. Wait, the $ G $ graph has a point at $ x = 3 $: looking at the grid, at $ x = 3 $, the $ G $ graph (light) has a $ y $-value. Wait, maybe the $ G $ graph at $ x = 3 $ is 6? Wait, no, let's check the $ G $ graph: the first segment (from $ x = 2 $ to $ x = 5 $ maybe) has a point at $ x = 3 $. Wait, actually, the $ G $ graph (light) at $ x = 3 $: looking at the graph, the $ G $ graph (the one with the open circle at $ x = 2 $, $ y = 6 $) and the closed dot at $ x = 2 $, $ y = 6 $? No, maybe the $ G $ graph at $ x = 3 $ is 6? Wait, no, let's re - examine. The $ G $ graph (light) has a point at $ x = 3 $: the $ y $-coordinate is 6? Wait, no, the $ G $ graph: the left part (before the peak) has a point at $ x = 2 $ (open circle, $ y = 6 $) and a closed dot at $ x = 2 $, $ y = 6 $? Wait, maybe the $ G $ graph at $ x = 3 $ is 6? Wait, no, let's think again. The $ G $ graph (light) has a peak at $ x = 5 $ (y = 8), then goes down. At $ x = 3 $, which is between $ x = 2 $ and $ x = 5 $, the line from $ x = 2 $ (y = 6) to $ x = 5 $ (y = 8). The slope is $ \frac{8 - 6}{5 - 2}=\frac{2}{3} $. So the equation is $ y - 6=\frac{2}{3}(x - 2) $. At $ x = 3 $, $ y=6+\frac{2}{3}(1)=6+\frac{2}{3}=\frac{20}{3}\approx6.67 $? But that doesn't seem right. Wait, maybe the $ G $ graph at $ x = 3 $ is 6? Wait, no, the graph shows that at $ x = 2 $, the $ G $ graph has a closed dot at $ y = 6 $ (maybe the open circle is a mistake, or I misread). Wait, maybe the $ G $ graph at $ x = 3 $ is 6? Wait, no, let's look at the grid. The $ G $ graph (light) at $ x = 3 $: the $ y $-axis has marks at 0, 2, 4, 6, 8, 10. At $ x = 3 $, the $ G $ graph is at $ y = 6 $? Wait, maybe the answer is 6. Wait, maybe the $ G $ graph at $ x = 3 $ is 6.
# Answer:
$ G(3)=6 $ (assuming the visual inspection gives $ y = 6 $ at $ x = 3 $ for $ G $)

### Question 19: Find $ F(4) $
# Explanation:
## Step 1: Locate $ x = 4 $ on the $ F $ graph (the dark - colored graph)
The $ F $ graph (dark) has a segment from $ x = 5 $ (or $ x = 4 $)? Wait, the $ F $ graph starts at $ x = 3 $ (or $ x = 4 $)? Wait, the $ F $ graph (dark) has a point at $ x = 5 $? No, let's see: the $ F $ graph (dark) goes from $ x = 3 $ (or $ x = 4 $)? Wait, the $ F $ graph (dark) has a point at $ x = 5 $? Wait, no, the $ F $ graph (dark) has a segment from $ x = 5 $ to $ x = 10 $? Wait, no, the $ F $ graph (dark) starts at $ x = 3 $ (the point at $ x = 5 $ is part of it). Wait, at $ x = 4 $, the $ F $ graph: the $ F $ graph (dark) has a line from $ x = 5 $ (y = 0) to $ x = 9 $ (y = 7) and then down. Wait, no, the $ F $ graph (dark) at $ x = 4 $: the $ F $ graph (dark) has a point at $ x = 5 $ (y = 0)? No, the $ F $ graph (dark) starts at $ x = 3 $ (the bottom point at $ x = 3 $, $ y=-2 $) and goes up to $ x = 5 $ (y = 0) and then to $ x = 9 $ (y = 7). Wait, the equation of the $ F $ graph from $ x = 3 $ to $ x = 9 $: slope is $ \frac{7 - (-2)}{9 - 3}=\frac{9}{6}=\frac{3}{2} $. So the equation is $ y+2=\frac{3}{2}(x - 3) $. At $ x = 4 $, $ y+2=\frac{3}{2}(1)$, so $ y=\frac{3}{2}-2=-\frac{1}{2} $? No, that can't be. Wait, maybe the $ F $ graph at $ x = 4 $ is 0? Wait, no, the $ F $ graph (dark) at $ x = 5 $ is 0 (the point at $ x = 5 $, $ y = 0 $). So at $ x = 4 $, which is between $ x = 3 $ and $ x = 5 $, using the slope: from $ x = 3 $ (y=-2) to $ x = 5 $ (y = 0), the change in x is 2, change in y is 2, so slope is 1. So equation: $ y+2 = 1\times(x - 3) $, so $ y=x - 5 $. At $ x = 4 $, $ y=4 - 5=-1 $? No, this is confusing. Wait, maybe the $ F $ graph at $ x = 4 $ is 0? Wait, no, the $ F $ graph (dark) has a point at $ x = 5 $ (y = 0), so at $ x = 4 $, maybe $ y = - 1 $? But this is unclear. Wait, maybe the $ F $ graph at $ x = 4 $ is 0? No, let's look at the grid again. The $ F $ graph (dark) at $ x = 4 $: the $ y $-axis has marks. Maybe the $ F $ graph at $ x = 4 $ is 0? Wait, I think I made a mistake. Wait, the $ F $ graph (dark) starts at $ x = 3 $ (the bottom point) and goes up to $ x = 5 $ (y = 0), so at $ x = 4 $, the $ y $-value is -1? But that seems odd. Wait, maybe the answer is 0? No, I think I need to re - evaluate. Wait, the $ F $ graph (dark) at $ x = 5 $ is 0, so at $ x = 4 $, which is one unit left of $ x = 5 $, and since the slope is 1 (from $ x = 3 $ (y=-2) to $ x = 5 $ (y = 0), slope is $ (0 - (-2))/(5 - 3)=1 $), so $ y=-2+(4 - 3)\times1=-1 $. So $ F(4)=-1 $? But this is not certain. Wait, maybe the $ F $ graph at $ x = 4 $ is 0. I think I need to check the graph again. Wait, the $ F $ graph (dark) has a point at $ x = 5 $ (y = 0), so at $ x = 4 $, it's -1.
# Answer:
$ F(4)=-1 $ (assuming the slope calculation is correct)

### Question 20: Find $ F(G(3)) $
# Explanation:
## Step 1: Find $ G(3) $
From question 18, we found $ G(3)=6 $ (assuming that).
## Step 2: Find $ F(6) $
Now we need to find $ F(6) $, i.e., the value of $ F $ at $ x = 6 $. The $ F $ graph (dark) at $ x = 6 $: using the equation of the $ F $ graph from $ x = 3 $ to $ x = 9 $ (slope 1, equation $ y=x - 5 $). At $ x = 6 $, $ y=6 - 5 = 1 $? Wait, no, the $ F $ graph (dark) at $ x = 6 $: looking at the graph, the $ F $ graph (dark) at $ x = 6 $ has a $ y $-value. Wait, the $ F $ graph (dark) goes from $ x = 3 $ (y=-2) to $ x = 9 $ (y = 7). At $ x = 6 $, which is 3 units from $ x = 3 $, so $ y=-2+3\times1 = 1 $. So $ F(6)=1 $. Wait, but let's check the graph. The $ F $ graph (dark) at $ x = 6 $: the $ y $-coordinate is 1? Or maybe 2? Wait, no, the equation $ y=x - 5 $ at $ x = 6 $ gives $ y = 1 $. So $ F(G(3))=F(6)=1 $.
# Answer:
$ F(G(3)) = 1 $ (assuming $ G(3)=6 $ and $ F(6)=1 $)

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QUESTION IMAGE

Question

Explanation:

Question 14: Find \( f(0) \)

Step 1: Locate \( x = 0 \) on the graph

Step 1: Locate \( x = -4 \) on the graph

Step 1: Locate \( x = 2 \) on the graph

Answer:

Question 15: Find \( f(-4) \)