Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, and origin at (0,0)

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 \(a>b\) and the major axis is along the \(y\)-axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b=2\).

Step2: Find the vertices and co - vertices

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

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 an ellipse passing through these four points. The ellipse will be 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 with \(a = 3\) (semi - major axis along \(y\)-axis) and \(b = 2\) (semi - minor axis along \(x\)-axis).
2. Mark the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\).
3. Sketch a smooth curve (ellipse) connecting these points, symmetric about both axes.

(Note: Since the question asks to graph the equation, the final answer is the graphical representation as described above. If we were to describe the key points for plotting: vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) with an ellipse passing through them.)

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (plotted and connected to form the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)).