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 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 vertical major axis ellipse (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1), the vertices are at ((0,\pm a)) and the co - vertices are at ((\pm b,0)).
  • Vertices: When (x = 0), from the equation (\frac{0^{2}}{4}+\frac{y^{2}}{9}=1), we get (y^{2}=9), so (y=\pm3). So the vertices are ((0, 3)) and ((0,- 3)).
  • Co - vertices: When (y = 0), from the equation (\frac{x^{2}}{4}+\frac{0^{2}}{9}=1), we get (x^{2}=4), so (x=\pm2). So the co - vertices are ((2,0)) and ((- 2,0)).

Step3: Plot the points and draw the ellipse

• Plot the vertices ((0,3)), ((0, - 3)) and co - vertices ((2,0)), ((-2,0)) on the coordinate plane.
• Then, draw a smooth ellipse passing through these four 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).

(Note: Since the question is about graphing, 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: )

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)). To graph it, plot these four points and draw a smooth curve connecting them, with the major axis along the (y) - axis and the minor axis along the (x) - axis.