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). Here, \(a=\sqrt{9}=3\) and \(b=\sqrt{4} = 2\).

Step2: Find the vertices and co - vertices

For a vertical 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 \((\pm b,0)\).

• Vertices: When \(x = 0\), \(y=\pm3\), so the vertices are \((0,3)\) and \((0, - 3)\).
• Co - vertices: When \(y = 0\), \(x=\pm2\), so the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and draw the ellipse

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. The major axis is along the \(y\) - axis with length \(2a=6\) and the minor axis is along the \(x\) - axis with length \(2b = 4\).

To graph the ellipse:

1. Locate the center of the ellipse. The center of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) is at the origin \((0,0)\) (since the equation is of the form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with \(h = 0\) and \(k=0\)).
2. Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the top and bottom most points of the ellipse).
3. Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the right - most and left - most points of the ellipse).
4. Draw a smooth, closed curve passing through these four points. The ellipse will be taller along the \(y\) - axis because \(a = 3\) (the semi - major axis length) is greater than \(b=2\) (the semi - minor axis length).

Answer:

The graph is an ellipse with center at \((0,0)\), vertices at \((0,\pm3)\), co - vertices at \((\pm2,0)\), and is drawn by connecting these points with a smooth curve. (The actual graph is a vertical ellipse passing through \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) as described above.)