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

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 a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the center is \((0,0)\) (since there are no shifts in \(x\) or \(y\) from the origin).

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

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane. Then, draw a smooth ellipse passing through these points. The major axis is along the \(y\) - axis with length \(2a = 6\) and the minor axis is along the \(x\) - axis with length \(2b=4\).

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\).)

Answer:

The graph is an ellipse centered at the origin \((0,0)\), with vertices at \((0, 3)\) and \((0, - 3)\), and co - vertices at \((2, 0)\) and \((-2, 0)\) (plotted and connected in a smooth elliptical curve).