Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines

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 of a vertical ellipse centered at the origin are \((0,\pm a)\), so the vertices are \((0, 3)\) and \((0,- 3)\).
• The co - vertices are \((\pm b,0)\), so the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then, sketch the ellipse passing through these points. The ellipse will be wider along the \(y\) - axis (since \(a>b\)) and symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it as a vertical ellipse centered at \((0,0)\) with \(a = 3\) (semi - major axis) and \(b=2\) (semi - minor axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse through these points, symmetric about the \(x\) - axis and \(y\) - axis. (The graph is an ellipse centered at the origin, with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\))