The problem asks us to construct the graphs of the following linear functions: a) $y = 2x - 1$ b) $y = 1 - x$ c) $y = -3x + 2$ d) $y = 4x$ e) $2x - 2y + 1 = 0$ f) $x - y + 3 = 0$ To draw the graph of a linear function, we need to find two points that lie on the line.

1. Problem Description

The problem asks us to construct the graphs of the following linear functions:

a) $y = 2x - 1$

b) $y = 1 - x$

c) $y = -3x + 2$

d) $y = 4x$

e) $2x - 2y + 1 = 0$

f) $x - y + 3 = 0$

To draw the graph of a linear function, we need to find two points that lie on the line.

2. Solution Steps

a) $y = 2x - 1$

If $x = 0$, then $y = 2(0) - 1 = -1$. So, the point $(0, -1)$ is on the line.

If $x = 1$, then $y = 2(1) - 1 = 1$. So, the point $(1, 1)$ is on the line.

b) $y = 1 - x$

If $x = 0$, then $y = 1 - 0 = 1$. So, the point $(0, 1)$ is on the line.

If $x = 1$, then $y = 1 - 1 = 0$. So, the point $(1, 0)$ is on the line.

c) $y = -3x + 2$

If $x = 0$, then $y = -3(0) + 2 = 2$. So, the point $(0, 2)$ is on the line.

If $x = 1$, then $y = -3(1) + 2 = -1$. So, the point $(1, -1)$ is on the line.

d) $y = 4x$

If $x = 0$, then $y = 4(0) = 0$. So, the point $(0, 0)$ is on the line.

If $x = 1$, then $y = 4(1) = 4$. So, the point $(1, 4)$ is on the line.

e) $2x - 2y + 1 = 0$

We can rewrite this equation as $2y = 2x + 1$, which means $y = x + \frac{1}{2}$.

If $x = 0$, then $y = 0 + \frac{1}{2} = \frac{1}{2}$. So, the point $(0, \frac{1}{2})$ is on the line.

If $x = 1$, then $y = 1 + \frac{1}{2} = \frac{3}{2}$. So, the point $(1, \frac{3}{2})$ is on the line.

f) $x - y + 3 = 0$

We can rewrite this equation as $y = x + 3$.

If $x = 0$, then $y = 0 + 3 = 3$. So, the point $(0, 3)$ is on the line.

If $x = 1$, then $y = 1 + 3 = 4$. So, the point $(1, 4)$ is on the line.

3. Final Answer

a) Two points on the line are $(0, -1)$ and $(1, 1)$.

b) Two points on the line are $(0, 1)$ and $(1, 0)$.

c) Two points on the line are $(0, 2)$ and $(1, -1)$.

d) Two points on the line are $(0, 0)$ and $(1, 4)$.

e) Two points on the line are $(0, \frac{1}{2})$ and $(1, \frac{3}{2})$.

f) Two points on the line are $(0, 3)$ and $(1, 4)$.