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

Explanation:

Step1: Identify the ellipse standard form

The given equation is (\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1), which is in the standard form of an ellipse (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1) (since (a>b) for vertical major axis), where (a^{2}=9) and (b^{2}=4). So, (a = 3) and (b = 2).

Step2: Find the vertices and co - vertices

For an ellipse with equation (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1), the vertices (endpoints of the major axis) are at ((0,\pm a)) and the co - vertices (endpoints of the minor axis) are at ((\pm b,0)). Substituting (a = 3) and (b = 2), the vertices are ((0, 3)) and ((0,- 3)), and the co - vertices are ((2,0)) and ((-2,0)).

Step3: Plot the points and draw the ellipse

Plot the points ((0,3)), ((0, - 3)), ((2,0)) and ((-2,0)) on the coordinate plane. Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be centered at the origin ((0,0)) with a vertical major axis (since the major axis is along the (y) - axis as (a) is associated with the (y) - term) and a horizontal minor axis.

To graph the ellipse:

• The center is ((0,0)).
• Move 3 units up and down from the center along the (y) - axis to get the vertices ((0,3)) and ((0, - 3)).
• Move 2 units left and right from the center along the (x) - axis to get the co - vertices ((2,0)) and ((-2,0)).
• Draw a smooth elliptical curve passing through these four points.

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

Answer:

The graph is an ellipse centered at the origin ((0,0)) with vertices ((0, 3)), ((0,-3)) and co - vertices ((2,0)), ((-2,0)) (represented by the smooth curve passing through these points).