Question

graph each equation.

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

Explanation:

Step1: Identify the ellipse equation 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 \(a > b\) for vertical major axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): 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 points \((0, 3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be centered at the origin \((0,0)\), with a vertical major axis (longer along the \(y\) - axis) passing through \((0,3)\) and \((0, - 3)\) and a horizontal minor axis passing through \((2,0)\) and \((-2,0)\).

(Note: Since the question is about graphing, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) centered at the origin.)

Answer:

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