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 center is \((0,0)\). The vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
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 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 and left most points of the ellipse).
• Then, draw a smooth curve connecting these points to form the ellipse.

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\). The ellipse is drawn by connecting these points with a smooth curve. (To actually draw it on the given grid, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and then sketch the ellipse passing through these points.)