The first step is to determine if the graph represents direct variation or partial variation. Since the graph passes through the point (0, 0), the graph represents direct variation. The equation for direct variation is y = mx, so the slope, m, must be determined.

Let (0, 0) = (x₁, y₁) and (20, 130) = (x₂, y₂).

Note: Point (10, 65) could be used; however, 65 is an estimation. It is always best to use a point where the exact values are known.

$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{130-0}{20-0}\\ & =& \frac{130}{20}\\ & =& 6.5\end{array}$

When m is substituted into the general equation y = mx, the equation of the line is y = 6.5x.

The first step is to determine if the graph represents direct variation or partial variation. This graph represents partial variation since the graph does not pass through the point (0, 0) but there is a y-intercept of 5. The equation for partial variation is y = mx + b, so the slope, m, and y-intercept, b, must be determined.

Let (0, 5) = (x₁, y₁) and (4, 10) = (x₂, y₂).

$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{10-5}{4-0}\\ & =& \frac{5}{4}\\ & =& 1.25\end{array}$

The y-intercept is 5 so b = 5.

Substitute m and b into the general equation y = mx + b. The equation of the line is y = 1.25x + 5.

The first step is to determine if the graph represents direct variation or partial variation.


The table of values represents direct variation since the graph passes through the point (0, 0) and the rate of change is constant (as the x-value increases by 3.8, the y-value increases by 20.9). The equation for direct variation is y = mx, so the slope, m, must be determined.

Let (0, 0) = (x₁, y₁) and (3.8, 20.9) = (x₂, y₂).

$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{20.9-0}{3.8-0}\\ & =& \frac{20.9}{3.8}\\ & =& 5.5\end{array}$

Substitute m into the general equation y = mx. The equation of the line is y = 5.5x.

The first step is to determine if the graph represents direct variation or partial variation.


The table of values represents partial variation since the graph does not pass through the point (0, 0) and the rate of change is constant (as the x-value increases by 3, the y-value increases by 21). The equation for partial variation is y = mx + b, so the slope, m, and y-intercept, b, must be determined.

Use the points (0, 8) and (3, 29) to find the slope.

Let (0, 8) = (x₁, y₁) and (3, 29) = (x₂, y₂).

$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{29-8}{3-0}\\ & =& \frac{21}{3}\\ & =& 7\end{array}$

The y-intercept of 8 is given by the point (0, 8). When m and b are substituted into the general equation y = mx + b, the equation of the line is y = 7x + 8.

1. n	C
   0	50
   1	65
   2	80
   3	95


The points are not connected since it is not possible to have part of a person (for example, 1.3 people). Therefore, that data is discrete.









2. n	C
   0	70
   1	85
   2	100
   3	115









3. Both graphs have the same slope, or rate of change of $15. The points in graph b are $20 higher than graph a since the initial cost (the fixed value, or y-intercept) is $20 higher.
   
   
4. Each point on the new graph will be $20 lower than the points on graph a and $40 lower than graph b.
   
   
5. The fixed value is the cost of the dinner if no one attended. Restaurants often charge a fixed amount to reserve an area for a large group.