Question

graph each equation.

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

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

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).

Step2: Find the values of \(a\) and \(b\)

For the given equation, \(a^2 = 9\) so \(a = 3\), and \(b^2=4\) so \(b = 2\).

Step3: Determine the vertices and co - vertices

• The vertices (endpoints of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (endpoints of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step4: Plot the points

• Plot the vertices \((0, 3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\).
• Then draw an ellipse passing through these four points.

To graph the ellipse:

1. Locate the points \((0,3)\), \((0, - 3)\), \((2,0)\), and \((-2,0)\) on the coordinate plane.
2. Sketch a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\)-axis (since \(a = 3\) and \(b=2\)) with its center at the origin \((0,0)\).

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, \pm 3)\) and co - vertices at \((\pm 2, 0)\), and the ellipse is drawn through these points on the given coordinate grid. (The actual drawing involves plotting the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketching the ellipse passing through them.)