Question

Hi, i need help with question 2! i need to graph all line equations for precalculus!

Answer 1

Let's make the graph of the following equation:

$\text{ 2x + 3y = 6}$

Before we start, let's first transform the given equation in the standard slope-intercept form: y = mx + b.

We get,

$\begin{gathered} \text{ 2x + 3y = 6} \\ \text{ 3y = 6 - 2x} \\ \text{ }\frac{\text{3y}}{\text{ 3}}\text{ = }\frac{\text{6 - 2x}}{\text{3}} \\ \text{ y = -}\frac{\text{ 2}}{\text{ 3}}x\text{ + 6} \end{gathered}$

Let's make the graph of the function.

To make the graph, let's identify at least two points that pass through its graph.

The easiest way is to find Point 1 at x = 0 and Point 2 at y = 0.

a.) Point 1 : x = 0

$\text{y = -}\frac{\text{ 2}}{\text{ 3}}x\text{ + 6}$$\text{ = -}\frac{\text{ 2}}{\text{ 3}}(0)\text{ + 6}$$\text{ = 0 + 6}$$\text{ y = 6}$

Thus, Point 1 : 0, 6

b.) Point 2 : y = 0

$\text{y = -}\frac{\text{ 2}}{\text{ 3}}x\text{ + 6}$$\text{0 = -}\frac{\text{ 2}}{\text{ 3}}x\text{ + 6}$$\frac{\text{ 2}}{\text{ 3}}x\text{ = 6}$$\text{ }x\text{ = 6(}\frac{\text{ 3}}{\text{ 2}})\text{ = }\frac{\text{ 18}}{\text{ 2}}$$x\text{ = 9}$

Therefore, Point 2 : 9, 0

Let's now plot the graph: