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 conic section type

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\) and the major axis is along the \(y\)-axis).

Step2: Determine the values of \(a\) and \(b\)

For the equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we have \(b^{2} = 4\) so \(b = 2\) and \(a^{2}=9\) so \(a = 3\).

Step3: Find the vertices and co - vertices

• The vertices of the ellipse (endpoints of the major axis, along the \(y\)-axis) are \((0,\pm a)=(0,\pm3)\).
• The co - vertices of the ellipse (endpoints of the minor axis, along the \(x\)-axis) are \((\pm b,0)=(\pm2,0)\).

Step4: Plot the points and draw the ellipse

1. Plot the vertices \((0, 3)\) and \((0, - 3)\) and the co - vertices \((2,0)\) and \((-2,0)\) on the coordinate plane.
2. Then, sketch a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\)-axis (since \(a = 3\) and \(b=2\)) and symmetric about both the \(x\)-axis and \(y\)-axis.

(Note: Since this is a graphing problem, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for plotting:

• Points on the \(x\)-axis: \((2,0)\) and \((-2,0)\)
• Points on the \(y\)-axis: \((0,3)\) and \((0, - 3)\)

And the ellipse is drawn through these points, symmetric about both axes.)

Answer:

Step1: Identify the conic section type

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\) and the major axis is along the \(y\)-axis).

Step2: Determine the values of \(a\) and \(b\)

For the equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we have \(b^{2} = 4\) so \(b = 2\) and \(a^{2}=9\) so \(a = 3\).

Step3: Find the vertices and co - vertices

• The vertices of the ellipse (endpoints of the major axis, along the \(y\)-axis) are \((0,\pm a)=(0,\pm3)\).
• The co - vertices of the ellipse (endpoints of the minor axis, along the \(x\)-axis) are \((\pm b,0)=(\pm2,0)\).

Step4: Plot the points and draw the ellipse

1. Plot the vertices \((0, 3)\) and \((0, - 3)\) and the co - vertices \((2,0)\) and \((-2,0)\) on the coordinate plane.
2. Then, sketch a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\)-axis (since \(a = 3\) and \(b=2\)) and symmetric about both the \(x\)-axis and \(y\)-axis.

(Note: Since this is a graphing problem, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for plotting:

• Points on the \(x\)-axis: \((2,0)\) and \((-2,0)\)
• Points on the \(y\)-axis: \((0,3)\) and \((0, - 3)\)

And the ellipse is drawn through these points, symmetric about both axes.)