Question

graph each equation.

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

coordinate plane grid with x from -8 to 8 and y from -8 to 8, axes labeled x and y

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 centered at the origin \((0,0)\)).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\):

• The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• The length of the semi - minor axis \(b = \sqrt{4}=2\), so the co - vertices are at \((\pm b,0)=(2,0)\) and \((- 2,0)\).

Step3: Plot the points

Plot the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane. Then, draw a smooth curve connecting these points to form the ellipse.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (the actual drawing involves plotting these points and connecting them with a smooth elliptical curve).