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 type

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 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 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 points on the y - axis, 3 units above and below the center).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the points on the x - axis, 2 units to the right and left of the center).
• Then, draw an ellipse passing through these four points, with the major axis along the y - axis and the minor axis along the x - axis.

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the vertices \((0,3)\) and \((0, - 3)\).
3. Mark the co - vertices \((2,0)\) and \((-2,0)\).
4. Sketch a smooth curve connecting these points, forming an ellipse that is taller along the y - axis (since the major axis is vertical) and narrower along the x - axis.

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

Answer:

Step1: Identify the conic section type

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 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 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 points on the y - axis, 3 units above and below the center).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the points on the x - axis, 2 units to the right and left of the center).
• Then, draw an ellipse passing through these four points, with the major axis along the y - axis and the minor axis along the x - axis.

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the vertices \((0,3)\) and \((0, - 3)\).
3. Mark the co - vertices \((2,0)\) and \((-2,0)\).
4. Sketch a smooth curve connecting these points, forming an ellipse that is taller along the y - axis (since the major axis is vertical) and narrower along the x - axis.

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