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

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 at ((0,0)). The vertices are at ((0,\pm a)) and the co - vertices are at ((\pm b,0)). Since (a=\sqrt{9} = 3) and (b=\sqrt{4}=2), the vertices are ((0, 3)) and ((0,- 3)), and the co - vertices are ((2,0)) and ((-2,0)).

Step3: Plot the points

Plot the center ((0,0)), the vertices ((0,3)), ((0, - 3)) and the co - vertices ((2,0)), ((-2,0)). Then draw the ellipse passing through these points. The major axis is along the (y) - axis with length (2a = 6) and the minor axis is along the (x) - axis with length (2b=4).

To graph the ellipse:

1. Mark the center at the origin ((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. Sketch a smooth curve connecting these points to form the ellipse.

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