Question

graph each equation.

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

coordinate grid with x from -8 to 8 and y from -8 to 8

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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2} = 4\), \(a>b\), so it is a vertical ellipse). Here, the center of the ellipse is at the origin \((0,0)\) because there are no shifts in \(x\) or \(y\) (the numerators are \(x^{2}\) and \(y^{2}\) without any linear terms).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the length of the semi - major axis \(a=\sqrt{9}=3\) and the length of the semi - minor axis \(b = \sqrt{4}=2\).

• The vertices (endpoints of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (endpoints of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

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

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 it is drawn by connecting these points with a smooth curve. (To actually draw it, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) on the given coordinate grid and sketch the ellipse through them.)