1. Change the equation into slope-intercept form.
The original equation is:
\[ 2x + 3y = 1,470 \]
To convert it into slope-intercept form (which is \( y = mx + b \)), we need to solve for \( y \).
- Subtract \( 2x \) from both sides:
\[ 3y = -2x + 1,470 \]
- Now, divide every term by \( 3 \):
\[ y = -\frac{2}{3}x + \frac{1,470}{3} \]
Calculating \( \frac{1,470}{3} \):
\[ \frac{1,470}{3} = 490 \]
So, the equation in slope-intercept form is:
\[ y = -\frac{2}{3}x + 490 \]
Identifying the slope and y-intercept:
- The slope (\( m \)) is \( -\frac{2}{3} \).
- The y-intercept (\( b \)) is \( 490 \).
2. Describe how you would graph this line using the slope-intercept method.
To graph the line using the slope-intercept method, you start with the y-intercept. From the equation \( y = -\frac{2}{3}x + 490 \), we know that the y-intercept is \( 490 \). This means that the point \( (0, 490) \) is where the line crosses the y-axis.
Next, we use the slope, which is \( -\frac{2}{3} \). This slope means that for every 3 units you move to the right along the x-axis, you move down 2 units along the y-axis. Starting from the y-intercept, I would plot the point \( (0, 490) \), then from this point, move 3 units right to \( x = 3 \) and 2 units down to \( y = 488 \). I would plot this second point at \( (3, 488) \).
Continuing this process, I could plot several points using the slope, and then draw a straight line through these points extending in both directions.
3. Write the equation in function notation. Explain what the graph of the function represents.
The equation in function notation can be written as:
\[ f(x) = -\frac{2}{3}x + 490 \]
This function \( f(x) \) represents the profit \( y \) Sal makes from selling \( x \) sandwiches, along with the profit coming from the wraps. The graph of this function shows the relationship between the number of sandwich lunch specials sold (the independent variable, \( x \)) and the total profit from those sandwiches and wraps (the dependent variable, \( f(x) \)). The slope indicates that as Sal sells more sandwiches, the profit decreases at a rate of \( \frac{2}{3} \) for each additional sandwich due to the fixed profit contributions from both sandwiches and wraps. The y-intercept indicates that if no sandwiches are sold (when \( x = 0 \)), the profit from wraps alone would amount to \( 490 \).