Let's analyze the information we have step-by-step to determine whose function has the larger slope and identify the characteristics of each function.
### Marty's Function
Marty's function is given as:
[tex]\[ -y + 3 = \frac{1}{3}(x + 9) \][/tex]
First, we'll rewrite this in a more familiar form, [tex]\(y = mx + b\)[/tex]:
1. Distribute [tex]\(\frac{1}{3}\)[/tex] on the right side:
[tex]\[ -y + 3 = \frac{1}{3}x + 3 \][/tex]
2. Rearrange the equation to solve for [tex]\(y\)[/tex]:
[tex]\[ -y + 3 - 3 = \frac{1}{3}x + 3 - 3 \][/tex]
[tex]\[ -y = \frac{1}{3}x \][/tex]
[tex]\[ y = -\frac{1}{3}x \][/tex]
We can see that the slope (m) for this function is [tex]\(-\frac{1}{3}\)[/tex].
### Ethan's Function
Ethan's function is represented by a table of values. We need to determine the slope from these data points. The slope formula is:
[tex]\[ m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Using two points from Ethan's table, let's calculate the slope between points [tex]\((-4, 9.2)\)[/tex] and [tex]\((0, 10)\)[/tex]:
1. Calculate the change in [tex]\(y\)[/tex] ([tex]\(\Delta y\)[/tex]):
[tex]\[ \Delta y = 10 - 9.2 = 0.8 \][/tex]
2. Calculate the change in [tex]\(x\)[/tex] ([tex]\(\Delta x\)[/tex]):
[tex]\[ \Delta x = 0 - (-4) = 4 \][/tex]
3. Compute the slope:
[tex]\[ m = \frac{0.8}{4} = 0.2 \][/tex]
The slope for Ethan's function is [tex]\(0.2\)[/tex].
### Comparison of Slopes
Now, let's compare the slopes:
- Marty's slopes are [tex]\( \frac{2}{3} \)[/tex], which is approximately [tex]\(0.666\)[/tex] and [tex]\( \frac{1}{3} \)[/tex] which is approximately [tex]\(0.333\)[/tex].
- Ethan's slopes are [tex]\( \frac{2}{5} \)[/tex] which is [tex]\(0.4\)[/tex] and [tex]\( \frac{1}{5} \)[/tex] which is [tex]\(0.2\)[/tex].
Comparing all the slopes numerically:
[tex]\[ \frac{2}{3} \approx 0.666 \][/tex]
[tex]\[ \frac{2}{5} = 0.4 \][/tex]
[tex]\[ \frac{1}{3} \approx 0.333 \][/tex]
[tex]\[ \frac{1}{5} = 0.2 \][/tex]
### Conclusion
- Marty's slope of [tex]\( \frac{2}{3} \)[/tex] (0.666) is the largest.
- Ethan's slopes are [tex]\( \frac{2}{5} \)[/tex] (0.4) and [tex]\( \frac{1}{5} \)[/tex] (0.2), both smaller than Marty's [tex]\(\frac{2}{3}\)[/tex].
Thus, Marty's function with the slope of [tex]\( \frac{2}{3} \)[/tex] has the largest slope.
