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

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 \(a^{2}=9\) and \(b^{2} = 4\), and \(a>b\), so it is a vertical ellipse). Here, \(a=\sqrt{9}=3\) and \(b=\sqrt{4} = 2\). The center of the ellipse is at the origin \((0,0)\) because there are no shifts in the \(x\) or \(y\) terms.

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).

• Vertices: Substituting \(a = 3\) into \((0,\pm a)\), we get the points \((0,3)\) and \((0, - 3)\).
• Co - vertices: Substituting \(b=2\) into \((\pm b,0)\), we get the points \((2,0)\) 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 \((2,0)\), \((-2,0)\) on the coordinate plane. Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the center \((0,0)\) on the coordinate grid.
2. Mark the vertices \((0,3)\) (3 units up from the center on the \(y\) - axis) and \((0, - 3)\) (3 units down from the center on the \(y\) - axis).
3. Mark the co - vertices \((2,0)\) (2 units to the right of the center on the \(x\) - axis) and \((-2,0)\) (2 units to the left of the center on the \(x\) - axis).
4. Draw a smooth, oval - shaped curve passing through these four points, making sure the curve is symmetric with respect to both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (the actual drawing involves plotting these points and connecting them with a smooth curve as described above).