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, and 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 \(a>b\), 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 at \((0,0)\). The vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
Since \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\), the vertices are \((0, 3)\) and \((0,- 3)\), and the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then draw 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 the origin \((0,0)\).
2. Move 3 units up and down from the center to get the vertices \((0,3)\) and \((0, - 3)\).
3. Move 2 units left and right from the center to get the co - vertices \((-2,0)\) and \((2,0)\).
4. Sketch a smooth curve connecting these points to form the ellipse.

(Note: Since the question asks to graph the equation, the final answer is the graph of the ellipse with center \((0,0)\), vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) as described above.)

Answer:

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

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then draw 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 the origin \((0,0)\).
2. Move 3 units up and down from the center to get the vertices \((0,3)\) and \((0, - 3)\).
3. Move 2 units left and right from the center to get the co - vertices \((-2,0)\) and \((2,0)\).
4. Sketch a smooth curve connecting these points to form the ellipse.

(Note: Since the question asks to graph the equation, the final answer is the graph of the ellipse with center \((0,0)\), vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) as described above.)