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

Explanation:

Step1: Identify the ellipse standard form

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 \(a>b\) for vertical major axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\).
• Then, sketch the ellipse by connecting these points smoothly, making sure it is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it is an ellipse with vertical major axis (since the denominator under \(y^{2}\) is larger).
2. Determine \(a = \sqrt{9}=3\) (distance from center to vertices on \(y\) - axis) and \(b=\sqrt{4} = 2\) (distance from center to co - vertices on \(x\) - axis).
3. Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((- 2,0)\).
4. Draw a smooth curve through these points, symmetric about the \(x\) - axis and \(y\) - axis.

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

Answer:

Step1: Identify the ellipse standard form

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 \(a>b\) for vertical major axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\).
• Then, sketch the ellipse by connecting these points smoothly, making sure it is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it is an ellipse with vertical major axis (since the denominator under \(y^{2}\) is larger).
2. Determine \(a = \sqrt{9}=3\) (distance from center to vertices on \(y\) - axis) and \(b=\sqrt{4} = 2\) (distance from center to co - vertices on \(x\) - axis).
3. Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((- 2,0)\).
4. Draw a smooth curve through these points, symmetric about the \(x\) - axis and \(y\) - axis.

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