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

Explanation:

Step1: Identify the ellipse standard form

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a > b\) for vertical major axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\).
• Then draw a smooth ellipse passing through these four points.

To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it is an ellipse with vertical major axis (since the denominator under \(y^{2}\) is larger).
2. Determine \(a = 3\) (distance from center to vertices on \(y\) - axis) and \(b = 2\) (distance from center to co - vertices on \(x\) - axis).
3. Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((- 2,0)\).
4. Draw a smooth curve connecting these points to form the ellipse.

(Note: Since the problem asks to graph, the key is to identify the type of conic (ellipse), find the key points (vertices and co - vertices) and plot them. The final graph will be an ellipse centered at the origin, with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\).)

Answer:

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