Question

graph each equation.

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

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 \(a > b>0\)) for an ellipse centered at the origin with a vertical major axis. Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

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

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0,- 3)\) and the co - vertices \((2,0)\), \((- 2,0)\).
• Then, sketch the ellipse by connecting these points smoothly. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

(Note: Since this is a graphing problem, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) centered at the origin. If we were to describe the key points for plotting:

• Vertices: \((0,3)\), \((0, - 3)\)
• Co - vertices: \((2,0)\), \((-2,0)\)

And the ellipse is drawn through these points with symmetry about the \(x\) and \(y\) axes.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\). (To draw it, plot these four points and sketch the ellipse connecting them, symmetric about both axes.)