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

Hi, i need help with question 2! i need to graph all line equations for precalculus-example-1

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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:

Hi, i need help with question 2! i need to graph all line equations for precalculus-example-1
User Dean Friedland
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