Question

graph each equation.

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

Explanation:

Step1: Identify the form of the equation

The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (\(a>b>0\)). Here, \(a^{2} = 9\), so \(a = 3\), and \(b^{2}=4\), so \(b = 2\).

Step2: Find the vertices and co - vertices

For an ellipse of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices are \((0,\pm a)\) and the co - vertices are \((\pm b,0)\).

• Vertices: When \(x = 0\), \(\frac{y^{2}}{9}=1\), so \(y=\pm3\). The vertices are \((0,3)\) and \((0, - 3)\).
• Co - vertices: When \(y = 0\), \(\frac{x^{2}}{4}=1\), so \(x=\pm2\). The co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Sketch the ellipse

Plot the vertices \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then draw a smooth curve passing through these four points to form the ellipse.

Answer:

The graph is an ellipse with vertices \((0,3)\) and \((0, - 3)\) and co - vertices \((2,0)\) and \((-2,0)\).