Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{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 \(a>b>0\)), which is a vertical - major - axis ellipse. Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For a vertical - major - axis ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
• Since \(a = 3\), the vertices are \((0,3)\) and \((0, - 3)\).
• Since \(b = 2\), the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and draw the ellipse

• Plot the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane.
• Then, sketch a smooth curve connecting these points to form the ellipse. The ellipse will be centered at the origin \((0,0)\) because the equation is of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) with no horizontal or vertical shifts (the center \((h,k)\) in the general form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) is \((0,0)\) here).

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the vertices at \((0,3)\) and \((0, - 3)\) (3 units up and down from the center along the y - axis).
3. Mark the co - vertices at \((2,0)\) and \((-2,0)\) (2 units left and right from the center along the x - axis).
4. Draw a smooth elliptical curve passing through these four points. The major axis is along the y - axis with length \(2a=6\) and the minor axis is along the x - axis with length \(2b = 4\).

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\). To draw it, plot the center, vertices, and co - vertices and sketch a smooth curve through them.