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

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's 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 points are \((0,3)\) and \((0, - 3)\).
• Co - vertices: When \(y = 0\), \(x=\pm2\), so the points 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 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 problem asks to graph the equation, the final answer is the graph of the ellipse with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) as described above. If we were to describe the graph in words, it's an ellipse centered at the origin, with a vertical major axis of length 6 and a horizontal minor axis of length 4.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\), with vertices at \((0, 3)\) and \((0, - 3)\), and co - vertices at \((2, 0)\) and \((-2, 0)\). To draw it, plot these four points and sketch a smooth elliptical curve through them.