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, origin at (0,0)

Explanation:

Step1: Identify the ellipse standard form

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (for a vertical major axis, where \(a>b\)). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\). The center of the ellipse is at \((0,0)\) (the origin) since there are no shifts in \(x\) or \(y\) in the equation.

Step2: Find the vertices and co - vertices

• For the vertical major axis ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (end - points of the major axis) are at \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are at \((\pm b,0)\).
• Vertices: When \(y=\pm a\) and \(x = 0\), so the vertices are \((0,3)\) and \((0, - 3)\).
• Co - vertices: When \(x=\pm b\) and \(y = 0\), 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 \((2,0)\), \((-2,0)\) on the coordinate plane.
• Then, sketch the ellipse by connecting these points smoothly. The ellipse will be taller along the \(y\) - axis (since \(a = 3\) and \(b = 2\)) with the center at the origin, passing through the vertices and co - vertices.

Answer:

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

1. Recognize it is an ellipse centered at \((0,0)\) with \(a = 3\) (vertical radius) and \(b=2\) (horizontal radius).
2. Plot vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse through these points. The ellipse is vertical (taller along \(y\) - axis) with center at the origin, passing through the plotted points.