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 \(a>b>0\)) for an ellipse centered at the origin with a vertical major axis. Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the major axis (vertical, along the \(y\) - axis), the vertices are at \((0,\pm a)=(0,\pm3)\).
• For the minor axis (horizontal, along the \(x\) - axis), the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\) and \((0, - 3)\) and the co - vertices \((2,0)\) and \((- 2,0)\).
• Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Recognize it is an ellipse centered at the origin \((0,0)\) with a vertical major axis (since the denominator under \(y^{2}\) is larger).
2. Determine the lengths of the semi - major axis \(a=\sqrt{9} = 3\) and semi - minor axis \(b=\sqrt{4}=2\).
3. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((- 2,0)\).
4. Draw a smooth curve through these points, symmetric about the \(x\) - axis and \(y\) - axis.

The graph will be an ellipse centered at the origin, with the top and bottom points (vertices) at \((0,3)\) and \((0, - 3)\), and the left and right points (co - vertices) at \((- 2,0)\) and \((2,0)\), and the curve of the ellipse passing through these four points.

Answer:

The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\), plotted and connected smoothly as described. (To present the final visual graph, you can plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((- 2,0)\) and draw the ellipse through them.)