Question

marty and ethan both wrote a function, but in different ways.
marty
$y + 3 = \frac{1}{3}(x + 9)$
ethan

x	-4	-2	0	2
y	9.2	9.6	10	10.4

whose function has the larger slope?
\bigcirc marty’s with a slope of $\frac{2}{3}$
\bigcirc ethan’s with a slope of $\frac{2}{5}$
\bigcirc marty’s with a slope of $\frac{1}{3}$
\bigcirc ethan’s with a slope of $\frac{1}{5}$

Explanation:

Step1: Find Marty's slope

Marty's equation: \( y + 3=\frac{1}{3}(x + 9) \). Rewrite in slope - intercept form \( y=mx + b \) (where \( m \) is the slope).
First, expand the right - hand side: \( y+3=\frac{1}{3}x + 3 \).
Then, subtract 3 from both sides: \( y=\frac{1}{3}x+3 - 3=\frac{1}{3}x \). So Marty's slope \( m_{Marty}=\frac{1}{3}\approx0.333 \).

Step2: Find Ethan's slope

Ethan's data is in a table. We can use the slope formula \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Let's take two points, say \( (x_1,y_1)=(- 4,9.2) \) and \( (x_2,y_2)=(-2,9.6) \).
\( m_{Ethan}=\frac{9.6 - 9.2}{-2-(-4)}=\frac{0.4}{2}=0.2 \)? Wait, no, let's check another pair. Take \( (x_1,y_1)=(0,10) \) and \( (x_2,y_2)=(2,10.4) \).
\( m_{Ethan}=\frac{10.4 - 10}{2 - 0}=\frac{0.4}{2}=0.2=\frac{1}{5} \)? Wait, no, wait, let's take \( (-2,9.6) \) and \( (0,10) \).
\( m=\frac{10 - 9.6}{0-(-2)}=\frac{0.4}{2}=0.2=\frac{1}{5} \). Wait, but wait, let's check \( (-4,9.2) \) and \( (0,10) \).
\( m=\frac{10 - 9.2}{0-(-4)}=\frac{0.8}{4}=0.2=\frac{1}{5} \). And Marty's slope is \( \frac{1}{3}\approx0.333 \), and \( \frac{1}{3}\approx0.333 \) and \( \frac{1}{5} = 0.2 \). Wait, no, maybe I made a mistake. Wait, let's re - calculate Marty's equation again. Wait, Marty's equation: \( y + 3=\frac{1}{3}(x + 9) \), so \( y=\frac{1}{3}x+3 - 3=\frac{1}{3}x \), slope is \( \frac{1}{3}\approx0.333 \). Ethan's slope: let's take \( (x_1,y_1)=(-2,9.6) \) and \( (x_2,y_2)=(2,10.4) \). Then \( m=\frac{10.4 - 9.6}{2-(-2)}=\frac{0.8}{4}=0.2=\frac{1}{5} \). Wait, but the options say "Ethan’s with a slope of 1/5" and "Marty’s with a slope of 1/3". Now, compare \( \frac{1}{3}\approx0.333 \) and \( \frac{1}{5}=0.2 \). Since \( \frac{1}{3}>\frac{1}{5} \), Marty's slope \( \frac{1}{3} \) is larger than Ethan's slope \( \frac{1}{5} \). Wait, but the options: "Marty’s with a slope of 1/3" is an option. Wait, let's re - check Marty's equation. The equation is \( y + 3=\frac{1}{3}(x + 9) \), so when we solve for \( y \), \( y=\frac{1}{3}x+3 - 3=\frac{1}{3}x \), so slope is \( \frac{1}{3} \). Ethan's slope: let's take two points from the table. Let's take \( (x=-4,y = 9.2) \) and \( (x = 0,y = 10) \). The change in \( x \) is \( 0-(-4)=4 \), change in \( y \) is \( 10 - 9.2 = 0.8 \). So slope \( m=\frac{0.8}{4}=0.2=\frac{1}{5} \). Now, \( \frac{1}{3}\approx0.333 \) and \( \frac{1}{5}=0.2 \). So Marty's slope \( \frac{1}{3} \) is larger than Ethan's slope \( \frac{1}{5} \).

Answer:

Marty’s with a slope of \( \frac{1}{3} \)