Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1\\)

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 \(a > b\) for a vertical major axis). Here, \(a^2=9\) so \(a = 3\), and \(b^2 = 4\) so \(b=2\). The center of the ellipse is at \((0,0)\) (the origin) since there are no shifts in the \(x\) or \(y\) terms.

Step2: Find the vertices and co - vertices

• For the major axis (along the \(y\) - axis, since \(a\) is associated with \(y^2\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the minor axis (along the \(x\) - axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points and draw the ellipse

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((-2,0)\).
• Then, draw a smooth ellipse connecting these points. The ellipse will be taller along the \(y\) - axis (since \(a = 3\)) and narrower along the \(x\) - axis (since \(b = 2\)).

(Note: Since the question is about graphing, 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. If we were to describe the key points for the graph: center \((0,0)\), 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 \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\), and a smooth curve connecting these points (taller along the \(y\) - axis and narrower along the \(x\) - axis).