Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1\\)

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, and origin at (0,0)

Explanation:

Step1: Identify the conic section

The equation \(\frac{x^2}{4}+\frac{y^2}{9} = 1\) is in the standard form of an ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (where \(a>b\) and the major axis is along the \(y\)-axis). Here, \(a^2 = 9\) so \(a = 3\), and \(b^2=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\)-intercepts (vertices, since major axis is along \(y\)-axis), set \(x = 0\). Then \(\frac{0^2}{4}+\frac{y^2}{9}=1\), so \(y^2=9\) and \(y=\pm3\). So the points are \((0, 3)\) and \((0, - 3)\).
• For the \(x\)-intercepts (co - vertices), set \(y = 0\). Then \(\frac{x^2}{4}+\frac{0^2}{9}=1\), so \(x^2 = 4\) and \(x=\pm2\). So the points are \((2, 0)\) and \((- 2, 0)\).

Step3: Plot the points

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then draw an ellipse passing through these points, centered at the origin \((0,0)\).

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, \pm3)\) and co - vertices at \((\pm2, 0)\), plotted as described above. (To actually draw it, connect the plotted points smoothly in an elliptical shape.)