Question

Which graph represents the linear equation y equals one half times x plus 2 on the coordinate plane?

graph of a line passing through the points 0 comma negative 2 and 2 comma negative 1
graph of a line passing through the points negative 4 comma 0 and 0 comma 2
graph of a line passing through the points negative 5 comma 0 and 0 comma 1
graph of a line passing through the points 0 comma 1 and negative 4 comma 0
Quest

Answer 1

To determine which graph represents the linear equation \( y = \frac{1}{2}x + 2 \), we can analyze the equation and find key points:

1. **Identify the y-intercept**: The equation is in slope-intercept form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Here, \( b = 2 \). This means that the line will cross the y-axis at the point (0, 2).

2. **Identify the slope**: The slope \( m = \frac{1}{2} \) indicates that for every 1 unit increase in \( x \), \( y \) increases by \( \frac{1}{2} \) units.

Using this information, we can find another point on the line, starting from the y-intercept (0, 2):

- If \( x = 2 \):
\[
y = \frac{1}{2}(2) + 2 = 1 + 2 = 3
\]
So, one of the points on the line is (2, 3).

Now let's analyze the given line options:

1. **Graph of line passing through (0, -2) and (2, -1)**: The y-intercept is -2 (not 2), so this graph does not represent our equation.

2. **Graph of line passing through (-4, 0) and (0, 2)**: This graph has the y-intercept at (0, 2) which matches our equation. To check the slope:
- From (-4, 0) to (0, 2):
\[
\text{slope} = \frac{2 - 0}{0 - (-4)} = \frac{2}{4} = \frac{1}{2}
\]
This matches our slope.

3. **Graph of line passing through (-5, 0) and (0, 1)**: The y-intercept is 1 (not 2), so this does not represent our equation.

4. **Graph passing through (0, 1) and (-4, 0)**: The y-intercept is 1 (not 2), so this graph does not represent our equation.

Based on this analysis, **the correct graph** that represents the equation \( y = \frac{1}{2}x + 2 \) is the **graph of a line passing through the points (-4, 0) and (0, 2)**.