1. **Stating the problem:** We need to graph the equations \(y = -3x\) and \(y = \frac{1}{3}x\). 2. **Formula and rules:** Both are linear equations in the form \(y = mx\), where \(m\) is the slope. - The slope \(m\) tells us how steep the line is. - A positive slope means the line rises as \(x\) increases. - A negative slope means the line falls as \(x\) increases. 3. **Graphing \(y = -3x\):** - When \(x=0\), \(y = -3 \times 0 = 0\). - When \(x=1\), \(y = -3 \times 1 = -3\). So points are \((0,0)\) and \((1,-3)\). 4. **Graphing \(y = \frac{1}{3}x\):** - When \(x=0\), \(y = \frac{1}{3} \times 0 = 0\). - When \(x=1\), \(y = \frac{1}{3} \times 1 = \frac{1}{3}\). So points are \((0,0)\) and \((1,\frac{1}{3})\). 5. **Explanation:** - The first line \(y = -3x\) is steep and goes downwards because the slope is \(-3\). - The second line \(y = \frac{1}{3}x\) is less steep and goes upwards because the slope is positive but less than 1. 6. **Final answer:** - Plot points \((0,0)\) and \((1,-3)\) for \(y = -3x\). - Plot points \((0,0)\) and \((1,\frac{1}{3})\) for \(y = \frac{1}{3}x\). - Draw straight lines through these points for each equation.