Question

graph each equation.

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

coordinate plane with x from -8 to 8 and y from -8 to 8, grid lines, axes labeled x and y

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 \(a>b>0\)) for an ellipse centered at the origin with a vertical major axis. Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the 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 \((\pm b,0)\).
• Vertices: When \(x = 0\), from the equation \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), we get \(y^{2}=9\), so \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• Co - vertices: When \(y = 0\), from the equation \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), we get \(x^{2}=4\), so \(x=\pm2\). So 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 co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane.
• Then, draw a smooth ellipse 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 \(2b = 4\).

(Note: Since the question is about graphing, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for graphing: )

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\). To graph it, plot these four points and draw a smooth curve connecting them, with the major axis along the \(y\) - axis and the minor axis along the \(x\) - axis.