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, origin at (0,0)

Explanation:

Step1: Identify the conic section type

The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a>b>0\)), so it is an ellipse centered at the origin \((0,0)\).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), we have \(a^{2}=9\) and \(b^{2}=4\). Taking the square roots, we get \(a = 3\) and \(b=2\).

• The vertices (end - points of the major axis, which is along the \(y\) - axis since \(a\) is under the \(y^{2}\) term) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis, along the \(x\) - 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 smoothly to form the ellipse. The ellipse will be taller along the \(y\) - axis (since the major axis is along the \(y\) - axis) with the vertices at \((0,3)\) and \((0, - 3)\) and co - vertices at \((2,0)\) and \((-2,0)\).

(Note: Since the question 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 the graph: vertices \((0, 3)\), \((0, - 3)\); co - vertices \((2, 0)\), \((-2, 0)\) and the ellipse passing through these points.)

Answer:

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