graph each equation.\n9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)\ngraph with x-axis from -8 to 8 and y-axis…

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^{2}=9) and (b^{2} = 4), and (a>b), so it is a vertical ellipse centered at the origin ((0,0))).

Step2: Find the vertices and co - vertices

For a vertical ellipse (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1):

• The value of (a=\sqrt{9}=3) and (b = \sqrt{4}=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 vertices ((0,3)) and ((0, - 3)).
• Plot the co - vertices ((2,0)) and ((-2,0)).
• Then, sketch the ellipse passing through these four points, making sure it is 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)) as described above. If we were to describe the key points for graphing: The ellipse is centered at the origin, goes up 3 units, down 3 units, left 2 units, and right 2 units from the center, and is a smooth curve connecting these extreme points.)

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)) (the actual graph is a smooth curve passing through these four points).