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 \(a > b\)), so it's 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 \(b^{2}=4\). Then \(a = 3\) and \(b=2\).
The vertices (end - points of the major axis, which is along the \(y\) - axis since \(a\) is under \(y^{2}\)) are \((0,\pm a)=(0,\pm3)\).
The co - vertices (end - points of the minor axis, along the \(x\) - axis) are \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points and draw the ellipse

• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((- 2,0)\).
• 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 \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the actual drawing on the grid connects these points in a smooth elliptical shape).