Question

graph each equation.

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

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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a > b>0\) and the major axis is along the \(y\)-axis). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

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 points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then, draw a smooth ellipse passing through these points. The ellipse will be taller along the \(y\)-axis (since \(a = 3\) is greater than \(b = 2\)) and symmetric about both the \(x\)-axis and \(y\)-axis.

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), plotted on the given coordinate grid and connected with a smooth curve. (To actually draw it, mark the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch the ellipse through them.)