Question

graph each equation.

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

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 \(a>b\) and major axis is along y - axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

• For the major axis (along y - axis), the vertices are \((0,\pm a)=(0,\pm3)\).
• For the minor axis (along x - axis), the co - vertices are \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the vertices \((0, 3)\), \((0,-3)\) and the co - vertices \((2,0)\), \((- 2,0)\).
• Then draw an ellipse passing through these four points. The ellipse will be symmetric about both the x - axis and y - axis.

(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. If we were to describe the key points for plotting:

• When \(x = 0\), \(y=\pm3\) (from \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\Rightarrow y^{2}=9\Rightarrow y = \pm3\)).
• When \(y = 0\), \(x=\pm2\) (from \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\Rightarrow x^{2}=4\Rightarrow x=\pm2\)).
• Then connect these points smoothly to form the ellipse.)

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 the ellipse is drawn passing through these four points (symmetric about x - axis and y - axis).