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 type

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)), so it is an ellipse centered at the origin ((0,0)).

Step2: Find the vertices and co - vertices

For the ellipse (\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1), we have (a^{2}=9) and (b^{2}=4). Taking the square roots, we get (a = 3) and (b=2).

• The vertices (end - points of the major axis, which is along the (y) - axis since (a) is under the (y^{2}) term) are at ((0,\pm a)=(0,\pm3)).
• The co - vertices (end - points of the minor axis, along the (x) - axis) are at ((\pm b,0)=(\pm2,0)).

Step3: Plot the points

Plot the points ((0,3)), ((0, - 3)), ((2,0)), and ((-2,0)) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be taller along the (y) - axis (since the major axis is along the (y) - axis) with the vertices at ((0,3)) and ((0, - 3)) and co - vertices at ((2,0)) and ((-2,0)).

(Note: Since the question asks to graph the equation, 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 the graph: vertices ((0, 3)), ((0, - 3)); co - vertices ((2, 0)), ((-2, 0)) and the ellipse passing through these points.)

Answer:

The graph is an ellipse centered at the origin with vertices at ((0, 3)), ((0, - 3)) and co - vertices at ((2, 0)), ((-2, 0)) (the ellipse is drawn by connecting these points smoothly).