Question

graph each equation. 9) \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\) graph with x from -8 to 8 and y 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 \((0,0)\). Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
• Vertices: When \(y=\pm a\), \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\)? Wait, no, \(a^{2}=9\) so \(a = 3\), so the vertices are \((0,3)\) and \((0,-3)\)? Wait, no, the standard form for vertical ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with center \((h,k)\). Here \(h = 0,k = 0\), \(a=\sqrt{9}=3\), \(b=\sqrt{4}=2\). So vertices are \((0,\pm a)=(0,3)\) and \((0, - 3)\)? Wait, no, \(a\) is the semi - major axis (along y - axis since \(a>b\)). So vertices are \((0,3)\) and \((0, - 3)\), co - vertices are \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((- 2,0)\).
• Then sketch the ellipse passing through these points. The ellipse will be taller along the y - axis (since the semi - major axis \(a = 3\) is along the y - axis) and have a width of \(2b=4\) along the x - axis and a height of \(2a = 6\) along the y - axis.

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

1. Identify the type of conic: The equation is of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (\(a = 3\), \(b = 2\), \(a>b\)), so it is a vertical ellipse centered at the origin \((0,0)\).
2. Find key points:

• Vertices (along the y - axis, since \(a\) is associated with \(y^{2}\)): \((0,\pm a)=(0,3)\) and \((0, - 3)\).
• Co - vertices (along the x - axis): \((\pm b,0)=(\pm2,0)\).

1. Plot the points: Mark the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((- 2,0)\) on the coordinate plane.
2. Sketch the ellipse: Draw a smooth curve connecting these points, making sure the ellipse is symmetric about both the x - axis and y - axis. The ellipse will be wider along the y - axis (from \(y=-3\) to \(y = 3\)) and have a width of \(4\) units along the x - axis (from \(x=-2\) to \(x = 2\)).

(Note: Since the question is to graph the equation, the final answer is the graph of the ellipse with center at \((0,0)\), vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above.)

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 is sketched by connecting these points with a smooth, symmetric curve.