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

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, making sure that the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Recognize it is an ellipse with center at \((0,0)\), vertical major axis (since \(a^{2}=9\) and \(b^{2}=4\), \(a > b\)), \(a = 3\), \(b = 2\).
2. Mark the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane.
3. Draw a smooth ellipse passing through these points, symmetric about the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\), sketched by plotting these points and drawing a smooth, symmetric curve through them. (The actual graph would have the ellipse passing through \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and curving smoothly between these points.)