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 \(a^{2}=9\) 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 \((0,0)\) (since there are no \(h\) and \(k\) shifts, \(h = 0,k=0\)).

• The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0, \pm a)=(0,\pm3)\).
• The length of the semi - minor axis \(b=\sqrt{4} = 2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((- 2,0)\).

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Recognize it is a vertical ellipse with center \((0,0)\), semi - major axis \(a = 3\) (along \(y\) - axis) and semi - minor axis \(b=2\) (along \(x\) - axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth curve through these points, symmetric about the \(x\) - axis and \(y\) - axis.

(Note: Since the question asks to graph the equation, the final answer is the graphical representation as described above. If we were to describe the key points for plotting: vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) with a smooth elliptical curve connecting them.)

Answer:

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 \((0,0)\) (since there are no \(h\) and \(k\) shifts, \(h = 0,k=0\)).

• The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0, \pm a)=(0,\pm3)\).
• The length of the semi - minor axis \(b=\sqrt{4} = 2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Plot the co - vertices \((2,0)\) and \((- 2,0)\).

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

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

1. Recognize it is a vertical ellipse with center \((0,0)\), semi - major axis \(a = 3\) (along \(y\) - axis) and semi - minor axis \(b=2\) (along \(x\) - axis).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth curve through these points, symmetric about the \(x\) - axis and \(y\) - axis.

(Note: Since the question asks to graph the equation, the final answer is the graphical representation as described above. If we were to describe the key points for plotting: vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) with a smooth elliptical curve connecting them.)