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

Step3: Plot the points

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

To graph the ellipse:

1. Mark the center at the origin \((0,0)\).
2. Move 3 units up and down from the center to mark the vertices \((0,3)\) and \((0, - 3)\).
3. Move 2 units left and right from the center to mark the co - vertices \((-2,0)\) and \((2,0)\).
4. Draw a smooth, closed curve passing through these four points. The ellipse will be taller along the \(y\) - axis (since the semi - major axis is along the \(y\) - axis) and narrower along the \(x\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, \pm 3)\) and co - vertices at \((\pm 2,0)\), drawn by plotting these points and connecting them with a smooth curve. (The actual graphing is done by marking the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and drawing a smooth ellipse through them on the given coordinate grid.)