Question

graph each equation.

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

coordinate plane with x from -8 to 8 and y from -8 to 8, grid lines

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 a vertical ellipse centered at the origin \((0,0)\):

• The vertices (endpoints of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (endpoints of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\) and \((0,- 3)\).
• Plot the co - vertices \((2,0)\) and \((-2,0)\).

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), drawn by connecting these points smoothly. (To actually draw it on the given grid, mark the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch the ellipse passing through these points, symmetric about both axes.)