What are the gradients of the linear graphs y=2x+1, y=2x, and y=2x−1 and how can they be plotted on a Cartesian plane?

The standard equation of a linear graph is given by:

where $m$ represents the gradient (slope) of the line and $c$ represents the y-intercept (the point where the line crosses the y-axis).

We compare each given equation to the standard form $y = mx + c$ to identify the gradient ($m$) and the y-intercept ($c$).

• For the equation $y = 2x + 1$: The gradient $m = 2$. The y-intercept $c = 1$.

• For the equation $y = 2x$: This can be written as $y = 2x + 0$. The gradient $m = 2$. The y-intercept $c = 0$.

• For the equation $y = 2x - 1$: The gradient $m = 2$. The y-intercept $c = -1$.

The completed table is:

From the completed table, the gradients of all three linear graphs are the same.
The gradients of the 3 linear graphs are $\boxed{\text{2}}$.

To plot each line, we can use the y-intercept and the gradient. The y-intercept is the point $(0, c)$. From this point, use the gradient $m = \frac{\text{rise}}{\text{run}} = \frac{2}{1}$ to find another point (move 1 unit right, 2 units up).

• For $y = 2x + 1$: Plot the y-intercept $(0, 1)$. From $(0, 1)$, move 1 unit right and 2 units up to reach $(1, 3)$. Draw a straight line through $(0, 1)$ and $(1, 3)$. (You can also find a third point, e.g., if $x=-1$, $y=2(-1)+1 = -1$, so $(-1, -1)$ is on the line).

• For $y = 2x$: Plot the y-intercept $(0, 0)$ (the origin). From $(0, 0)$, move 1 unit right and 2 units up to reach $(1, 2)$. Draw a straight line through $(0, 0)$ and $(1, 2)$. (You can also find a third point, e.g., if $x=-1$, $y=2(-1) = -2$, so $(-1, -2)$ is on the line).

• For $y = 2x - 1$: Plot the y-intercept $(0, -1)$. From $(0, -1)$, move 1 unit right and 2 units up to reach $(1, 1)$. Draw a straight line through $(0, -1)$ and $(1, 1)$. (You can also find a third point, e.g., if $x=-1$, $y=2(-1)-1 = -3$, so $(-1, -3)$ is on the line).

When plotted, these three lines will be parallel to each other because they all have the same gradient ($m=2$).