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 \(b^{2} = 4\), 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 \((0,0)\). The vertices (end - points of the major axis) are \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)\).
Given \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\)? Wait, no. Wait, \(a^{2}=9\), so \(a = 3\), and \(b^{2}=4\), so \(b = 2\). Wait, the standard form of a vertical ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with center \((h,k)\). Here \(h = 0,k = 0\), \(a=\sqrt{9}=3\), \(b=\sqrt{4}=2\). So the vertices (on the major axis, \(y\) - axis) are \((0, k\pm a)=(0,3)\) and \((0, - 3)\)? No, wait, major axis is along the \(y\) - axis when \(a>b\), so the vertices are \((0, a)\) and \((0, - a)\), so \((0,3)\) and \((0, - 3)\), and the co - vertices (on the minor axis, \(x\) - axis) are \((b,0)\) and \((-b,0)\), so \((2,0)\) and \((-2,0)\).

Step3: Plot the center, vertices, and co - vertices

• The center of the ellipse is at the origin \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the points on the \(y\) - axis, 3 units above and below the center).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the points on the \(x\) - axis, 2 units to the right and left of the center).

Step4: Sketch the ellipse

Using the center, vertices, and co - vertices, sketch the ellipse by drawing a smooth curve that passes through the vertices and co - vertices. 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. Identify the type of conic: It is an ellipse with a vertical major axis (since the denominator of \(y^{2}\) is larger than the denominator of \(x^{2}\)).
2. Find key points:

• Center: \((0,0)\) (the origin).
• Vertices (major axis endpoints): \((0, \pm 3)\) (since \(a=\sqrt{9}=3\), along the \(y\) - axis).
• Co - vertices (minor axis endpoints): \((\pm 2,0)\) (since \(b=\sqrt{4}=2\), along the \(x\) - axis).

1. Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), and \((-2,0)\).
2. Draw the ellipse by connecting these points with a smooth, oval - shaped curve, symmetric about both the \(x\) - axis and \(y\) - axis.

(Note: Since the question is about graphing, the final answer is the graph as described above. If we were to describe the graph in words, it is an ellipse centered at the origin, with vertices at \((0, 3)\) and \((0, - 3)\) and co - vertices at \((2, 0)\) and \((-2, 0)\), symmetric about the \(x\) - and \(y\) - axes.)

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 \((0,0)\). The vertices (end - points of the major axis) are \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)\).
Given \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\)? Wait, no. Wait, \(a^{2}=9\), so \(a = 3\), and \(b^{2}=4\), so \(b = 2\). Wait, the standard form of a vertical ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with center \((h,k)\). Here \(h = 0,k = 0\), \(a=\sqrt{9}=3\), \(b=\sqrt{4}=2\). So the vertices (on the major axis, \(y\) - axis) are \((0, k\pm a)=(0,3)\) and \((0, - 3)\)? No, wait, major axis is along the \(y\) - axis when \(a>b\), so the vertices are \((0, a)\) and \((0, - a)\), so \((0,3)\) and \((0, - 3)\), and the co - vertices (on the minor axis, \(x\) - axis) are \((b,0)\) and \((-b,0)\), so \((2,0)\) and \((-2,0)\).

Step3: Plot the center, vertices, and co - vertices

• The center of the ellipse is at the origin \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the points on the \(y\) - axis, 3 units above and below the center).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the points on the \(x\) - axis, 2 units to the right and left of the center).

Step4: Sketch the ellipse

Using the center, vertices, and co - vertices, sketch the ellipse by drawing a smooth curve that passes through the vertices and co - vertices. 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. Identify the type of conic: It is an ellipse with a vertical major axis (since the denominator of \(y^{2}\) is larger than the denominator of \(x^{2}\)).
2. Find key points:

• Center: \((0,0)\) (the origin).
• Vertices (major axis endpoints): \((0, \pm 3)\) (since \(a=\sqrt{9}=3\), along the \(y\) - axis).
• Co - vertices (minor axis endpoints): \((\pm 2,0)\) (since \(b=\sqrt{4}=2\), along the \(x\) - axis).

1. Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), and \((-2,0)\).
2. Draw the ellipse by connecting these points with a smooth, oval - shaped curve, symmetric about both the \(x\) - axis and \(y\) - axis.

(Note: Since the question is about graphing, the final answer is the graph as described above. If we were to describe the graph in words, it is an ellipse centered at the origin, with vertices at \((0, 3)\) and \((0, - 3)\) and co - vertices at \((2, 0)\) and \((-2, 0)\), symmetric about the \(x\) - and \(y\) - axes.)