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

Step2: Determine \(a\) and \(b\)

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

Step3: Find the vertices and co - vertices

• Vertices: For a vertical ellipse centered at the origin, the vertices are at \((0,\pm a)\). So the vertices are \((0, 3)\) and \((0,-3)\).
• Co - vertices: The co - vertices are at \((\pm b,0)\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

Step4: Plot the points and draw the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then, sketch the ellipse passing through these points. The ellipse will be taller along the \(y\) - axis (since \(a = 3\) is along the \(y\) - axis) and wider along the \(x\) - axis by \(b = 2\) units from the center.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (plotted on the given coordinate grid as described in the steps). To draw it, plot the center \((0,0)\), then the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch a smooth curve connecting these points to form the ellipse.