Question

graph each equation.

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

coordinate grid with x from -8 to 8 and y from -8 to 8

Explanation:

Step1: Identify the ellipse standard form

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 \(a > b\), major axis is vertical). Here, \(a=\sqrt{9} = 3\), \(b=\sqrt{4}=2\), and the center is at \((0,0)\).

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 and draw the ellipse

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), and \((- 2,0)\) on the coordinate plane. Then, sketch a smooth ellipse passing through these points. The ellipse will be taller along the \(y\) - axis (since the major axis is vertical with length \(2a = 6\)) and wider along the \(x\) - axis with length \(2b=4\).

Answer:

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

1. Recognize it as an ellipse with center \((0,0)\), \(a = 3\) (semi - major axis, vertical), \(b = 2\) (semi - minor axis, horizontal).
2. Plot vertices \((0,3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse through these points, taller along the \(y\) - axis and wider along the \(x\) - axis. (The actual graph is an ellipse centered at the origin, with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\))