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 matches the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (for a vertical major axis, where \(a > b>0\)). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is associated with \(y^{2}\)): 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, sketch the ellipse by connecting these points smoothly, making sure the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Recognize it is an ellipse with a vertical major axis (since the denominator under \(y^{2}\) is larger).
2. The vertices are \((0, 3)\) and \((0,-3)\), and the co - vertices are \((2, 0)\) and \((-2, 0)\).
3. Plot these four points and draw a smooth, symmetric ellipse passing through them.

(Note: Since the question 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 key points for graphing: vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) and the ellipse passing through these points.)

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2, 0)\), \((-2, 0)\), symmetric about the \(x\) - axis and \(y\) - axis. (To draw it, plot these points and sketch a smooth ellipse through them.)