Question

graph each equation.

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

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 vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
Given \(a^{2}=9\), so \(a = 3\) (we take the positive square root since \(a\) represents a length). Given \(b^{2}=4\), so \(b = 2\) (we take the positive square root since \(b\) represents a length).
So the vertices are \((0,3)\) and \((0, - 3)\), and 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.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. 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 question is to graph the equation, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for the graph: The ellipse is centered at the origin \((0,0)\), has vertices at \((0,3)\) and \((0, - 3)\), and co - vertices at \((2,0)\) and \((-2,0)\).)

Answer:

The graph is an ellipse centered at the origin, with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\), and is drawn by connecting these points smoothly.