Question

graph each equation.

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

coordinate plane with x from -8 to 8 and y from -8 to 8, grid lines

Explanation:

Step1: Identify the ellipse standard 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 an ellipse with equation \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (endpoints of the major axis) are at \((0,\pm a)\) and the co - vertices (endpoints of the minor axis) are at \((\pm b,0)\).
Substituting \(a = 3\) and \(b = 2\), the vertices are \((0, 3)\) and \((0,- 3)\), and the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and draw the ellipse

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be centered at the origin \((0,0)\) with a vertical major axis (since the major axis is along the \(y\) - axis as \(a\) is associated with the \(y\) - term) and a horizontal minor axis.

To graph the ellipse:

• The center is \((0,0)\).
• Move 3 units up and down from the center along the \(y\) - axis to get the vertices \((0,3)\) and \((0, - 3)\).
• Move 2 units left and right from the center along the \(x\) - axis to get the co - vertices \((2,0)\) and \((-2,0)\).
• Draw a smooth elliptical curve passing through these four points.

(Note: Since the question asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices \((0, 3)\), \((0,-3)\) and co - vertices \((2,0)\), \((-2,0)\) (represented by the smooth curve passing through these points).