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, origin at (0,0)

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 \(a^2 = 9\) and \(b^2=4\), and \(a>b\), so it is a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), the center is at \((h,k)=(0,0)\) (since there are no shifts in \(x\) and \(y\) from the origin).

• The length of the semi - major axis \(a=\sqrt{9} = 3\), so the vertices are at \((0,\pm a)=(0,\pm3)\).
• The length of the semi - minor axis \(b=\sqrt{4}=2\), so 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 centered at the origin, with a vertical major axis (longer axis) of length \(2a = 6\) (from \(y=-3\) to \(y = 3\)) and a horizontal minor axis (shorter axis) of length \(2b=4\) (from \(x=-2\) to \(x = 2\)).

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the vertices \((0,3)\) and \((0, - 3)\) on the \(y\) - axis.
3. Mark the co - vertices \((2,0)\) and \((-2,0)\) on the \(x\) - axis.
4. Draw a smooth curve connecting these points, making sure the curve is symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), and the ellipse is drawn by connecting these points smoothly. (The actual drawing would be an ellipse as described above on the given coordinate grid.)