graph each equation.\n9) \\(\\dfrac{x^2}{4} + \\dfrac{y^2}{9} = 1\\)\ncoordinate grid with x 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) (since (a^2 = 9) and (b^2=4), and (a>b), so it is a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse (\frac{x^2}{b^2}+\frac{y^2}{a^2}=1), the center is ((0,0)) (since there are no shifts in (x) or (y) from the origin).

• The length of the semi - major axis (a=\sqrt{9} = 3), so the vertices are at ((0, a)=(0,3)) and ((0, - a)=(0, - 3)).
• The length of the semi - minor axis (b=\sqrt{4}=2), so the co - vertices are at ((b,0)=(2,0)) and ((-b,0)=(-2,0)).

Step3: Plot the points and draw the ellipse

Plot the center ((0,0)), the vertices ((0,3)), ((0, - 3)) and the co - vertices ((2,0)), ((-2,0)). Then, sketch the ellipse passing through these points, making sure it is symmetric about both the (x) - axis and (y) - axis.

To graph the ellipse:

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

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at ((0,0)), vertices at ((0,\pm3)) and co - vertices at ((\pm2,0)) as described above.)