Question

Graphing lines. Plot the following lines:

A. y=3+2xy=3+2x
B. y=1+xy=1+x
C. y=−2+3xy=−2+3x
D. y=5xy=5x
E. y=4−2xy=4−2x

Answer 1

To graph a line of the form $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept.

• Start by plotting the y-intercept which is (0, $b$).
• Find another point using the slope $m$ with the y-intercept at the reference point.
• Connect the two points.

A. To graph the line $y=3+2x$, we know that the slope is 2 and the y-intercept is (0, 3).

The fact that the slope of the line is 2 tells us that when x increases by 1 unit, y increases by 2 units. We know this because of the formula of the slope $m=\frac{\Delta y}{\Delta x}$.

Therefore, another point with the y-intercept at the reference point is (0 + 1, 3 + 2) = (1, 5).

Connecting this points, we have the following graph.

B. To graph the line $y=1+x$, we know that the slope is 1 and the y-intercept is (0, 1).

Another point with the y-intercept at the reference point is (0 + 1, 1 + 1) = (1, 2).

Connecting this points, we have the following graph.

C. To graph the line $y=-2+3x$, we know that the slope is 3 and the y-intercept is (0, -2).

Another point with the y-intercept at the reference point is (0 + 1, -2 + 3) = (1, 1).

Connecting this points, we have the following graph.

D. To graph the line $y=5x$, we know that the slope is 5 and the y-intercept is (0, 0).

Another point with the y-intercept at the reference point is (0 + 1, 0 + 5) = (1, 5).

Connecting this points, we have the following graph.

E. To graph the line $y=4-2x$, we know that the slope is -2 and the y-intercept is (0, 4).

Another point with the y-intercept at the reference point is (0 + 1, 4 - 2) = (1, 2).

Connecting this points, we have the following graph.