Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, and origin at (0,0)

Explanation:

Step1: Identify the type of conic

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

Step2: Find the vertices and co - vertices

For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (end - points of the major axis) are \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)\).
Given \(a^{2}=9\), so \(a = 3\) (we take the positive root since \(a\) represents a length), and \(b^{2}=4\), so \(b = 2\) (we take the positive root since \(b\) represents a length).
So the vertices are \((0,3)\) and \((0, - 3)\), and the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points

• Plot the vertices \((0,3)\) and \((0,-3)\) on the \(y\) - axis.
• Plot the co - vertices \((2,0)\) and \((-2,0)\) on the \(x\) - axis.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be centered at the origin \((0,0)\) (since the equation is of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) with no horizontal or vertical shifts), with the major axis along the \(y\) - axis (length \(2a = 6\)) and the minor axis along the \(x\) - axis (length \(2b=4\)).

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, \pm 3)\) and co - vertices at \((\pm 2,0)\), drawn by connecting these points smoothly. (The actual drawing would show an ellipse that is taller than it is wide, passing through \((0,3)\), \((0, - 3)\), \((2,0)\), and \((-2,0)\) and curving between these points.)