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

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 \(a>b\) and the major axis is along the \(y\)-axis). Here, \(a^2 = 9\) so \(a = 3\) and \(b^2=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\)-axis (major axis): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\)-axis (minor axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0, - 3)\), \((2,0)\) and \((- 2,0)\).
• Then draw an ellipse passing through these points, 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 center \((h,k)=(0,0)\)).

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. From the center, move 3 units up and down along the \(y\)-axis to get the vertices \((0,3)\) and \((0, - 3)\).
3. From the center, move 2 units left and right along the \(x\)-axis to get the co - vertices \((2,0)\) and \((-2,0)\).
4. Sketch a smooth ellipse passing through these four points.

(Note: Since the problem 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. If we were to describe the key points for the graph: )

Answer:

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