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 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 \(b^{2}=4\), so \(a = 3\), \(b = 2\)) with a vertical major axis (because \(a^{2}>b^{2}\) and the \(y\)-term has the larger denominator).

Step2: Find the vertices and co - vertices

• The vertices of the ellipse (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 points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points with a smooth curve to form the ellipse. The ellipse will be centered at the origin \((0,0)\) (since there are no shifts in the \(x\) or \(y\) terms in the equation), with the major axis along the \(y\) - axis and the minor axis along the \(x\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), and the ellipse is drawn by connecting these points with a smooth curve. (The actual drawing would be done on the provided coordinate grid by plotting the points and sketching the ellipse through them.)