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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2}=4\), and \(a > b\)), so it is an ellipse centered at the origin \((0,0)\) with a vertical major axis.

Step2: Find the vertices and co - vertices

For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (\(a>b>0\)):

• The length of the semi - major axis \(a=\sqrt{9} = 3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• The length of the semi - minor axis \(b=\sqrt{4}=2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and it is drawn by connecting these points smoothly. (To actually draw it, plot the center, vertices, and co - vertices and sketch the ellipse passing through them.)