Question

graph each equation.

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

coordinate grid with x from -8 to 8 and y from -8 to 8

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 \(a > b>0\)), which is a vertical ellipse centered at the origin \((0,0)\).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), we have \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

• The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis) 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, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Move 3 units up and down from the center to get the vertices \((0,3)\) and \((0, - 3)\).
3. Move 2 units left and right from the center to get the co - vertices \((2,0)\) and \((-2,0)\).
4. Draw a smooth curve connecting these four points, forming an ellipse that is wider along the \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), sketched by connecting the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) with a smooth curve. (The actual graph is drawn on the given coordinate grid with the above - mentioned key points.)