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, 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).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the center is \((0,0)\). The vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
Since \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\), the vertices are \((0, 3)\) and \((0,- 3)\), and the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\) 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 it: Start by marking the center at the origin. Then go up 3 units from the center to \((0,3)\) and down 3 units to \((0, - 3)\) for the vertices. Go right 2 units from the center to \((2,0)\) and left 2 units to \((-2,0)\) for the co - vertices. Then sketch the ellipse passing through these four 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 actual drawing should be a smooth curve connecting these points as described in the steps).