Question

graph each equation.

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

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 \(a>b\) and the \(y\)-term has the larger denominator). The center of the ellipse is at the origin \((0,0)\) because there are no shifts in the \(x\) or \(y\) terms.

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: Since \(a = 3\), the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: Since \(b = 2\), the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then, sketch the ellipse by connecting these points smoothly, making sure that the ellipse is symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Locate the center at \((0,0)\).
2. Mark the vertices on the \(y\) - axis: \((0,3)\) and \((0, - 3)\).
3. Mark the co - vertices on the \(x\) - axis: \((2,0)\) and \((-2,0)\).
4. Draw a smooth ellipse passing through these four points, symmetric about the \(x\) and \(y\) axes.

(Note: Since the problem asks to graph the equation, the final answer is the graphical representation as described above. If we were to describe the key points for the graph: center \((0,0)\), vertices \((0,\pm3)\), co - vertices \((\pm2,0)\) and the ellipse drawn through them.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\), drawn as a smooth curve passing through these points, symmetric about the \(x\) - axis and \(y\) - axis.