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 \(a>b>0\)), which is a vertical 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 \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

• The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\) and \((0, - 3)\) on the \(y\) - axis.
• Plot the co - vertices \((2,0)\) and \((- 2,0)\) on the \(x\) - axis.

Step4: Draw the ellipse

Connect the plotted points \((0,3)\), \((2,0)\), \((0, - 3)\), and \((- 2,0)\) with a smooth curve to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

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 is a smooth curve passing through these points and symmetric about both axes).