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 is \(\frac{x^2}{4}+\frac{y^2}{9} = 1\), which is the standard form of an ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (since \(a^2 = 9\) and \(b^2=4\), so \(a = 3\), \(b = 2\)) with a vertical major axis (because \(a>b\) and the \(y^2\) term has the larger denominator). The center is at \((0,0)\) (the origin).

Step2: Find the vertices and co - vertices

• For the vertices (endpoints of the major axis): Since the major axis is vertical, the vertices are at \((0,\pm a)=(0,\pm3)\).
• For the co - vertices (endpoints of the minor axis): Since the minor axis is horizontal, the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((- 2,0)\).

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

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

1. Recognize it as an ellipse with center \((0,0)\), vertical major axis (since \(9>4\)), \(a = 3\) (distance from center to vertices on \(y\) - axis), \(b = 2\) (distance from center to co - vertices on \(x\) - axis).
2. Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\), and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse passing through these points, symmetric about the \(x\) - axis and \(y\) - axis. The ellipse has a vertical major axis with vertices at \((0,\pm3)\) and horizontal minor axis with co - vertices at \((\pm2,0)\).