Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)

graph grid with x from -8 to 8 and y from -8 to 8

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).

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

For the given equation, \(a^2 = 9\) so \(a=\sqrt{9}=3\), and \(b^2 = 4\) so \(b=\sqrt{4}=2\).

Step3: Determine the vertices and co - vertices

• The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step4: Plot the points

• Plot the points \((0, 3)\), \((0,-3)\), \((2,0)\) and \((- 2,0)\) on the coordinate plane.
• Then, draw a smooth curve connecting these points to form the ellipse.

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

1. Recognize it is an ellipse with major axis along the \(y\) - axis, \(a = 3\) (distance from center to vertices on \(y\) - axis) and \(b=2\) (distance from center to co - vertices on \(x\) - axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth elliptical curve passing through these points.

(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 \((2,0)\), \((-2,0)\). The ellipse is drawn by connecting these points with a smooth curve.