The first step is to determine if the graph represents direct variation or partial variation. Since the graph does not pass through the point (0, 0) and there is a y-intercept of –2, this graph represents partial variation. The equation for partial variation is y = mx + b, so the slope, or rate of change, m, and y-intercept, b, must be determined.

Let (1, 1) = (x₁, y₁) and (2, 4) = (x₂, y₂)

$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{4-1}{2-1}\\ & =& \frac{3}{1}\\ & =& 3\end{array}$

The y-intercept is –2 and this is the b-value. Substitute m and b into the general equation y = mx + b. The equation of the line is y = 3x – 2.

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, or rate of change, m, must be determined.

Let (0, 0) = (x₁, y₁) and (5, 7) = (x₂, y₂).

$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{7-0}{5-0}\\ & =& \frac{7}{5}\\ & =& 1.4\end{array}$

Substitute m into the general equation y = mx. The equation of the line is y = 1.4x.

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 the rate of change is constant (as the x-value increases by 2, the y-value increases by 16). The equation for partial variation is y = mx + b, so the slope, m, and y-intercept, b, must be determined.

Let (0, 4.1) = (x₁, y₁) and (2, 20.1) = (x₂, y₂).
 
$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{20.1-4.1}{2-0}\\ & =& \frac{16}{2}\\ & =& 8\\ & & \end{array}$
The y-intercept of 4.1 is given by the point (0, 4.1). Therefore, b = 4.1. Substitute m and b into the general equation y = mx + b. The equation of the line is y = 8x + 4.1.

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


The graph 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 6, the y-value increases by 1.56). The equation for direct variation is y = mx, so the slope, m, must be determined.

Let (0, 0) = (x₁, y₁) and (6, 1.56) = (x₂, y₂).
 
$\begin{array}{rcl}m& =& \frac{y_2-y_1}{x_2-x_1}\\ & =& \frac{1.56-0}{6-0}\\ & =& \frac{1.56}{6}\\ & =& 0.26\end{array}$

When m is substituted into the general equation y = mx, the equation of the line is y = 0.26x.