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, axes labeled x and y

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 \(b^{2} = 4\), and \(a>b\), so it is a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the center is at \((0,0)\). The vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
Given \(a^{2}=9\), so \(a = 3\) (we take the positive square root since \(a\) represents a length), and \(b^{2}=4\), so \(b = 2\) (we take the positive square root since \(b\) represents a length).
So 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 center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the top and bottom most points of the ellipse).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the right - most and left - most points of the ellipse).
• Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

(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. If we were to describe the key points for graphing: center \((0,0)\), 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 \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\) (the ellipse is drawn by connecting these points smoothly).