Question

match the graphs to their equations. f
\frac { ( x + 3 ) ^ { 2 } } { 9 } + \frac { ( y + 1 ) ^ { 2 } } { 16 } = 1
\frac { ( x + 3 ) ^ { 2 } } { 16 } + \frac { ( y - 1 ) ^ { 2 } } { 9 } = 1
\frac { ( x - 3 ) ^ { 2 } } { 16 } + \frac { ( y + 1 ) ^ { 2 } } { 9 } = 1
\frac { ( x - 3 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1

Explanation:

To solve the problem of matching the ellipse equations to their graphs, we use the standard form of an ellipse: \(\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1\) (vertical major axis) or \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\) (horizontal major axis), where \((h, k)\) is the center, \(a\) is the semi - major axis, and \(b\) is the semi - minor axis.

1. Analyze the equation \(\frac{(x + 3)^2}{9}+\frac{(y + 1)^2}{16}=1\)

• The center of the ellipse is \((h,k)=(- 3,-1)\) (since \(x+3=x - (-3)\) and \(y + 1=y-(-1)\)).
• The denominator under the \(y\) - term (\(16\)) is larger than the denominator under the \(x\) - term (\(9\)), so the major axis is vertical.
• The semi - major axis \(a=\sqrt{16} = 4\) and the semi - minor axis \(b=\sqrt{9}=3\).
• Looking at the graphs, graph b is centered at \(x\approx - 3,y\approx - 1\) (from the grid) and has a vertical major axis. So \(\frac{(x + 3)^2}{9}+\frac{(y + 1)^2}{16}=1\) matches graph b.

2. Analyze the equation \(\frac{(x + 3)^2}{16}+\frac{(y - 1)^2}{9}=1\)

• The center of the ellipse is \((h,k)=(-3,1)\).
• The denominator under the \(x\) - term (\(16\)) is larger than the denominator under the \(y\) - term (\(9\)), so the major axis is horizontal.
• The semi - major axis \(a = \sqrt{16}=4\) and the semi - minor axis \(b=\sqrt{9} = 3\).
• None of the given graphs (a, b) seem to match this center \((-3,1)\) with a horizontal major axis. (We assume there are more graphs not fully shown, but based on the given ones, we can still analyze the other equations)

3. Analyze the equation \(\frac{(x - 3)^2}{16}+\frac{(y + 1)^2}{9}=1\)

• The center of the ellipse is \((h,k)=(3,-1)\).
• The denominator under the \(x\) - term (\(16\)) is larger than the denominator under the \(y\) - term (\(9\)), so the major axis is horizontal.
• The semi - major axis \(a=\sqrt{16} = 4\) and the semi - minor axis \(b=\sqrt{9}=3\).
• Graph a is centered at \(x\approx3,y\approx - 1\) (from the grid) and has a horizontal major axis. So \(\frac{(x - 3)^2}{16}+\frac{(y + 1)^2}{9}=1\) matches graph a.

4. Analyze the equation \(\frac{(x - 3)^2}{9}+\frac{(y - 1)^2}{16}=1\)

• The center of the ellipse is \((h,k)=(3,1)\).
• The denominator under the \(y\) - term (\(16\)) is larger than the denominator under the \(x\) - term (\(9\)), so the major axis is vertical.
• The semi - major axis \(a=\sqrt{16}=4\) and the semi - minor axis \(b=\sqrt{9}=3\).
• None of the given graphs (a, b) seem to match this center \((3,1)\) with a vertical major axis. (Again, assuming more graphs are present)

Final Matches (for the given graphs a and b)

• \(\frac{(x + 3)^2}{9}+\frac{(y + 1)^2}{16}=1\) \(

ightarrow\) Graph b

• \(\frac{(x - 3)^2}{16}+\frac{(y + 1)^2}{9}=1\) \(

ightarrow\) Graph a

If we consider the first equation \(\frac{(x + 3)^2}{9}+\frac{(y + 1)^2}{16}=1\) and graph b:

Answer:

\(\frac{(x + 3)^2}{9}+\frac{(y + 1)^2}{16}=1\) matches graph b; \(\frac{(x - 3)^2}{16}+\frac{(y + 1)^2}{9}=1\) matches graph a. (For the other two equations, more graph information is needed for a complete match)