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, axes labeled x and y

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 \(a>b\), 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 at \((0,0)\). 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 and draw the ellipse

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the top and bottom most points of the ellipse).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the right - most and left - most points of the ellipse).
• Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it as a vertical ellipse with center \((0,0)\).
2. Vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\).
3. Plot the center, vertices, and co - vertices, then draw a smooth elliptical curve through them, symmetric about the \(x\) - axis and \(y\) - axis. (The actual graph is an ellipse centered at the origin, with a vertical major axis of length \(6\) (from \(y=- 3\) to \(y = 3\)) and a horizontal minor axis of length \(4\) (from \(x=-2\) to \(x = 2\)).)