Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)

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 \(a > b>0\) and the major axis is along the \(y\)-axis). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\)-intercepts (vertices, since major axis is along \(y\)-axis), set \(x = 0\). Then \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), which gives \(y^{2}=9\), so \(y=\pm3\). The points are \((0, 3)\) and \((0,- 3)\).
• For the \(x\)-intercepts (co - vertices), set \(y = 0\). Then \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), which gives \(x^{2}=4\), so \(x=\pm2\). The points are \((2,0)\) and \((- 2,0)\).

Step3: Sketch the ellipse

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane and draw a smooth curve connecting these points to form the ellipse.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. But in text form, we can describe the key points and the shape.)

Answer:

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