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 \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (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 \(a = 3\) (distance from center \((0,0)\) to vertices on \(y\) - axis) and \(b = 2\) (distance from center to co - vertices on \(x\) - axis).
3. Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\).
4. Draw a smooth curve through these points, symmetric about both axes.

(Note: Since the question 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.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), drawn by connecting these points smoothly.