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 \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (where \(a > b>0\)), which is a vertical ellipse centered at the origin \((0,0)\).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), we have \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

• 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 \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0,-3)\), \((2,0)\) and \((- 2,0)\) on the coordinate plane.
• Then, draw a smooth ellipse passing through these four points. The major axis is along the \(y\) - axis (vertical) with length \(2a = 6\) and the minor axis is along the \(x\) - axis (horizontal) with length \(2b=4\).

(Note: Since this is a graphing problem, 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 the graph: the ellipse is centered at the origin, has a vertical major axis with vertices at \((0, 3)\) and \((0,-3)\), and horizontal minor axis with co - vertices at \((2,0)\) and \((-2,0)\), and the graph is a smooth curve connecting these points.)

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 through these four points with the major axis vertical and minor axis horizontal).