Question

graph each equation. 9) \\( \frac{x^2}{4} + \frac{y^2}{9} = 1 \\) graph with x from -8 to 8, 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 \(a>b>0\)), so it is an ellipse centered at the origin \((0,0)\).

Step2: Find the vertices and co - vertices

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

• The vertices (end - points of the major axis, which is along the \(y\) - axis since \(a\) is under the \(y^{2}\) term) are \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis, along the \(x\) - axis) are \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.
• Then, sketch the ellipse by connecting these points smoothly. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\), and it is sketched by connecting these points smoothly on the given coordinate grid.