Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, origin at (0,0)

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 \(a>b\), the major axis is along the \(y\)-axis).

Step2: Find the vertices and co - vertices

For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (end - points of the major axis) are \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)\).
Given \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\).
Given \(b^{2}=4\), then \(b = 2\), so the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points

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

To graph the ellipse:

1. Locate the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the given coordinate grid.
2. Draw a smooth, closed curve passing through these points. The ellipse will be taller along the \(y\) - axis (since the major axis is along the \(y\) - axis with length \(2a=6\)) and wider along the \(x\) - axis with length \(2b = 4\).

Answer:

The graph is an ellipse with vertices at \((0,3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the actual drawing involves plotting these points and drawing a smooth curve through them).