Question

based on the table list if the function is an example of direct variation.
x:2,4,6,8
y:36,52,68,84
If the function is a direct variation name the constant and if not name the slope
write the equation in slope intercept form

Answer 1

To determine if the relationship between \( x \) and \( y \) is a direct variation, we need to analyze the data points provided. A direct variation can be represented by the equation \( y = kx \), where \( k \) is the constant of variation. For direct variation, the ratio \( \frac{y}{x} \) should be constant for all pairs of \( x \) and \( y \).

Let’s calculate \( \frac{y}{x} \) for each pair:

1. For \( x = 2, y = 36 \):
\[
\frac{y}{x} = \frac{36}{2} = 18
\]

2. For \( x = 4, y = 52 \):
\[
\frac{y}{x} = \frac{52}{4} = 13
\]

3. For \( x = 6, y = 68 \):
\[
\frac{y}{x} = \frac{68}{6} \approx 11.33
\]

4. For \( x = 8, y = 84 \):
\[
\frac{y}{x} = \frac{84}{8} = 10.5
\]

The ratios are not constant (\( 18, 13, 11.33, 10.5 \)), indicating that the relationship between \( x \) and \( y \) is **not** a direct variation.

Next, we can determine the slope of the relationship using the pairs of points. The slope \( m \) can be calculated using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Let's choose two points to calculate the slope, for example (2, 36) and (4, 52):

\[
m = \frac{52 - 36}{4 - 2} = \frac{16}{2} = 8
\]

Now, using the slope we found, we can write the equation in slope-intercept form \( y = mx + b \). To find \( b \), we can substitute one of the points into the equation. Let's use the point \( (2, 36) \):

\[
36 = 8(2) + b
\]
\[
36 = 16 + b
\]
\[
b = 36 - 16 = 20
\]

So the equation in slope-intercept form is:

\[
y = 8x + 20
\]

### Summary
- The function is **not** an example of direct variation.
- The slope \( m \) is \( 8 \).
- The equation in slope-intercept form is \( y = 8x + 20 \).