Table.pdf - FOA/Algebra 1 Unit 5: Comparing Linear Quadratic and ... - Free Printable

Let’s solve this step by step.

We are given two functions:

- Function f(x) is shown as a graph (a straight line going down from left to right).
- Function g(x) is given in a table:

| x | g(x) |
|----|------|
| -2 | -10 |
| -1 | -5 |
| 0 | -2 |
| 1 | -1 |
| 2 | -2 |

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Part a: Compare the y-intercepts of the two functions.

The y-intercept is where the graph crosses the y-axis — that’s when x = 0.

For f(x): Look at the graph. When x = 0, the line is at y = 4. So f(0) = 4.

For g(x): Look at the table. When x = 0, g(0) = -2.

So:
- f(x) has y-intercept = 4
- g(x) has y-intercept = -2

→ f(x) has a greater y-intercept than g(x).

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Part b: Compare the slopes of the two functions.

Slope tells us how steep the line is and whether it goes up or down.

For f(x): It’s a straight line. Let’s pick two points on the graph to calculate slope.

From the graph:
- At x = 0, y = 4 → point (0, 4)
- At x = 4, y = 0 → point (4, 0)

Slope = (change in y) / (change in x) = (0 - 4) / (4 - 0) = (-4)/4 = -1

So slope of f(x) = -1

For g(x): We need to check if it’s linear first. Let’s look at the differences between consecutive outputs.

From x = -2 to x = -1: g changes from -10 to -5 → change = +5
From x = -1 to x = 0: g changes from -5 to -2 → change = +3
From x = 0 to x = 1: g changes from -2 to -1 → change = +1
From x = 1 to x = 2: g changes from -1 to -2 → change = -1

The changes are not constant → so g(x) is NOT linear, meaning it doesn’t have a single slope.

But the question says “compare the slopes”. Since g(x) isn’t linear, we can’t assign one slope to it. However, maybe they want us to compare the *average* rate of change over some interval? Or perhaps just note that f(x) has a constant slope while g(x) does not?

Looking back at the problem: “Compare the slopes of the two functions.”

Since g(x) is not linear, it doesn’t have a single slope. But let’s see what the average rate of change is from x = -2 to x = 2 for g(x):

Change in g(x): from -10 to -2 → Δy = 8
Change in x: from -2 to 2 → Δx = 4
Average slope = 8/4 = 2

But that’s an average — not the actual slope everywhere.

Alternatively, maybe the question expects us to realize that since g(x) is curved (not straight), its slope changes, while f(x) has a fixed slope of -1.

But let’s check the values again — actually, looking at the table:

Wait — let me double-check the table:

x | g(x)
--|-----
-2 | -10
-1 | -5
0 | -2
1 | -1
2 | -2

From x=-2 to x=-1: rise = 5, run=1 → slope=5
x=-1 to 0: rise=3, run=1 → slope=3
x=0 to 1: rise=1, run=1 → slope=1
x=1 to 2: rise=-1, run=1 → slope=-1

So the slope of g(x) is changing — decreasing each time.

Whereas f(x) has a constant slope of -1.

So comparison:

→ f(x) has a constant negative slope of -1.
→ g(x) has varying slopes: positive at first, then becomes negative. Its slope decreases as x increases.

But perhaps for simplicity, since the question might expect a basic comparison, we can say:

“f(x) has a constant slope of -1. g(x) is not linear, so its slope changes — it starts positive and becomes negative.”

But let’s see part c — maybe that will help clarify.

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Part c: Compare the values of g(x) for the two functions.

Wait — this seems misworded. It says: “Compare the values of g(x) for the two functions.” That doesn’t make sense — g(x) is only one function.

Probably it means: “Compare the values of the two functions” — i.e., compare f(x) and g(x) at certain x-values.

Let’s assume that’s what it means.

Let’s pick a few x-values and compare f(x) and g(x).

First, we need to know f(x) at those x-values. From the graph:

At x = -2: f(-2) = ? Let’s estimate from graph.

Graph of f(x): passes through (0,4) and (4,0). Equation: since slope is -1, equation is y = -x + 4.

Check: at x=0, y=4; x=4, y=0 → yes.

So f(x) = -x + 4

Now compute f(x) and g(x) at same x-values:

At x = -2:
f(-2) = -(-2) + 4 = 2 + 4 = 6
g(-2) = -10
→ f(-2) > g(-2)

At x = -1:
f(-1) = -(-1) + 4 = 1 + 4 = 5
g(-1) = -5
→ f(-1) > g(-1)

At x = 0:
f(0) = 4
g(0) = -2
→ f(0) > g(0)

At x = 1:
f(1) = -1 + 4 = 3
g(1) = -1
→ f(1) > g(1)

At x = 2:
f(2) = -2 + 4 = 2
g(2) = -2
→ f(2) > g(2)

So at all these x-values, f(x) > g(x)

Is there any x where g(x) > f(x)? Let’s try x=3:

f(3) = -3 + 4 = 1
g(3) — not given, but from pattern, after x=2, g(x) was -2 at x=2, and before that it was increasing then decreasing. Probably continues decreasing? Not sure.

But based on given data, for all x from -2 to 2, f(x) is always greater than g(x).

So answer: For the x-values given, f(x) is always greater than g(x).

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Part d: Create a table of values for the function h(x) = f(x) + g(x)

We already have f(x) = -x + 4

And g(x) from table.

So h(x) = f(x) + g(x) = (-x + 4) + g(x)

Let’s compute for x = -2, -1, 0, 1, 2

At x = -2:
f(-2) = 6, g(-2) = -10 → h(-2) = 6 + (-10) = -4

At x = -1:
f(-1) = 5, g(-1) = -5 → h(-1) = 5 + (-5) = 0

At x = 0:
f(0) = 4, g(0) = -2 → h(0) = 4 + (-2) = 2

At x = 1:
f(1) = 3, g(1) = -1 → h(1) = 3 + (-1) = 2

At x = 2:
f(2) = 2, g(2) = -2 → h(2) = 2 + (-2) = 0

So table for h(x):

| x | h(x) |
|----|------|
| -2 | -4 |
| -1 | 0 |
| 0 | 2 |
| 1 | 2 |
| 2 | 0 |

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Now, let’s summarize answers clearly.

Final Answer:

a. The y-intercept of f(x) is 4, and the y-intercept of g(x) is -2. So f(x) has a greater y-intercept.

b. f(x) has a constant slope of -1. g(x) is not linear, so its slope changes: it is positive for x < 0 and negative for x > 1, decreasing overall.

c. For all x-values from -2 to 2, f(x) is greater than g(x).

d. Table for h(x) = f(x) + g(x):

| x | h(x) |
|----|------|
| -2 | -4 |
| -1 | 0 |
| 0 | 2 |
| 1 | 2 |
| 2 | 0 |