Question

graph each equation.

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

coordinate plane with x from -8 to 8 and y from -8 to 8, grid lines, axes labeled x and y

Explanation:

Step1: Identify the conic section

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which 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 centered at the origin \((0,0)\)).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\):

• The value of \(a=\sqrt{9}=3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• The value of \(b = \sqrt{4}=2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane. Then, sketch the 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 \(2b=4\).

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the vertices: move 3 units up from the center to \((0,3)\) and 3 units down to \((0, - 3)\).
3. Mark the co - vertices: move 2 units to the right from the center to \((2,0)\) and 2 units to the left to \((-2,0)\).
4. Draw a smooth curve connecting these points to form the ellipse.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (the actual graph is a smooth curve passing through these points as described in the steps).