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, 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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a>b\) for vertical major axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b=2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: Since the major axis is along the \(y\) - axis, the vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor axis) co - vertices: The co - vertices are at \((\pm b,0)=(\pm2,0)\).

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 four points. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it is an ellipse with \(a = 3\) (semi - major axis along \(y\) - axis) and \(b = 2\) (semi - minor axis along \(x\) - axis).
2. Plot vertices \((0,3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw an ellipse through these points, symmetric about \(x\) and \(y\) axes. The graph is an ellipse centered at the origin \((0,0)\), with top vertex at \((0,3)\), bottom vertex at \((0, - 3)\), right co - vertex at \((2,0)\) and left co - vertex at \((-2,0)\).