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

The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\) and the major axis is along the \(y\)-axis). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the major axis (along \(y\)-axis), the vertices are at \((0,\pm a)=(0,\pm3)\).
• For the minor axis (along \(x\)-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 that the ellipse 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 major axis along \(y\)-axis, \(a = 3\), \(b=2\).
2. Plot vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth curve through these points, symmetric about \(x\) and \(y\) axes.

(Note: Since the problem asks to graph the equation, 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 the graph: vertices \((0, 3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\))

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), symmetric about the \(x\) - axis and \(y\) - axis. (The actual graph is a smooth curve passing through these points.)