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 ellipse standard form

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

Step2: Find the vertices and co - vertices

• For a vertical ellipse (\(a\) is under \(y^{2}\)), the vertices are at \((0,\pm a)=(0,\pm3)\) and the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points and draw the ellipse

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((-2,0)\).
• Then, draw a smooth ellipse passing through these four points. The major axis is along the \(y\) - axis (length \(2a = 6\)) and the minor axis is along the \(x\) - axis (length \(2b=4\)).

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), drawn by connecting these points smoothly. (The actual drawing would be on the provided coordinate grid with the above - mentioned points plotted and the ellipse traced through them.)