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 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 \(a>b\) 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 ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), the vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\) and the co - vertices (end - points of the minor axis) 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 curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\) - axis since the major axis is along the \(y\) - axis (because \(a = 3\) and \(b = 2\), and \(a>b\)).

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the actual graph is a smooth curve passing through these points as described in the steps).