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### Hyperbola
The equation of the hyperbola is given by:
[tex]\[ y = \frac{k}{x} + q \][/tex]
The hyperbola passes through the points [tex]\((-9, -\frac{1}{3})\)[/tex] and [tex]\((3, 1)\)[/tex].
1. Substitute the point [tex]\((-9, -\frac{1}{3})\)[/tex] into the equation:
[tex]\[ -\frac{1}{3} = \frac{k}{-9} + q \][/tex]
Simplifying, we get:
[tex]\[ -\frac{1}{3} = -\frac{k}{9} + q \][/tex]
2. Substitute the point [tex]\((3, 1)\)[/tex] into the equation:
[tex]\[ 1 = \frac{k}{3} + q \][/tex]
Now we have a system of linear equations:
[tex]\[ -\frac{1}{3} = -\frac{k}{9} + q \][/tex]
[tex]\[ 1 = \frac{k}{3} + q \][/tex]
3. Solve the system of equations:
- Multiply the first equation by 9 to clear the fraction:
[tex]\[ 9 \left(-\frac{1}{3}\right) = 9 \left(-\frac{k}{9} + q\right) \][/tex]
[tex]\[ -3 = -k + 9q \][/tex]
[tex]\[ k = 9q + 3 \][/tex]
- Substitute [tex]\( k = 9q + 3 \)[/tex] into the second equation:
[tex]\[ 1 = \frac{9q + 3}{3} + q \][/tex]
[tex]\[ 1 = 3q + 1 + q \][/tex]
[tex]\[ 1 = 4q + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ 0 = 4q \][/tex]
[tex]\[ q = 0 \][/tex]
- Substitute [tex]\( q = 0 \)[/tex] back into [tex]\( k = 9q + 3 \)[/tex]:
[tex]\[ k = 9(0) + 3 \][/tex]
[tex]\[ k = 3 \][/tex]
Therefore, the values are:
[tex]\[ k = 3 \][/tex]
[tex]\[ q = 0 \][/tex]
### Parabola
The equation of the parabola is given by:
[tex]\[ y = ax^2 + q \][/tex]
The parabola has a turning point at [tex]\((0, 3)\)[/tex]. This tells us that when [tex]\( x = 0 \)[/tex]:
[tex]\[ y = q \][/tex]
So, [tex]\( q = 3 \)[/tex].
The parabola also passes through the point [tex]\((3, 12)\)[/tex].
1. Substitute the point [tex]\((3, 12)\)[/tex] into the equation:
[tex]\[ 12 = a(3)^2 + 3 \][/tex]
Simplifying, we get:
[tex]\[ 12 = 9a + 3 \][/tex]
Subtract 3 from both sides:
[tex]\[ 9 = 9a \][/tex]
[tex]\[ a = 1 \][/tex]
Therefore, the values are:
[tex]\[ a = 1 \][/tex]
[tex]\[ q = 3 \][/tex]
### Summary
For the hyperbola:
[tex]\[ k = 3 \][/tex]
[tex]\[ q = 0 \][/tex]
For the parabola:
[tex]\[ a = 1 \][/tex]
[tex]\[ q = 3 \][/tex]
### Hyperbola
The equation of the hyperbola is given by:
[tex]\[ y = \frac{k}{x} + q \][/tex]
The hyperbola passes through the points [tex]\((-9, -\frac{1}{3})\)[/tex] and [tex]\((3, 1)\)[/tex].
1. Substitute the point [tex]\((-9, -\frac{1}{3})\)[/tex] into the equation:
[tex]\[ -\frac{1}{3} = \frac{k}{-9} + q \][/tex]
Simplifying, we get:
[tex]\[ -\frac{1}{3} = -\frac{k}{9} + q \][/tex]
2. Substitute the point [tex]\((3, 1)\)[/tex] into the equation:
[tex]\[ 1 = \frac{k}{3} + q \][/tex]
Now we have a system of linear equations:
[tex]\[ -\frac{1}{3} = -\frac{k}{9} + q \][/tex]
[tex]\[ 1 = \frac{k}{3} + q \][/tex]
3. Solve the system of equations:
- Multiply the first equation by 9 to clear the fraction:
[tex]\[ 9 \left(-\frac{1}{3}\right) = 9 \left(-\frac{k}{9} + q\right) \][/tex]
[tex]\[ -3 = -k + 9q \][/tex]
[tex]\[ k = 9q + 3 \][/tex]
- Substitute [tex]\( k = 9q + 3 \)[/tex] into the second equation:
[tex]\[ 1 = \frac{9q + 3}{3} + q \][/tex]
[tex]\[ 1 = 3q + 1 + q \][/tex]
[tex]\[ 1 = 4q + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ 0 = 4q \][/tex]
[tex]\[ q = 0 \][/tex]
- Substitute [tex]\( q = 0 \)[/tex] back into [tex]\( k = 9q + 3 \)[/tex]:
[tex]\[ k = 9(0) + 3 \][/tex]
[tex]\[ k = 3 \][/tex]
Therefore, the values are:
[tex]\[ k = 3 \][/tex]
[tex]\[ q = 0 \][/tex]
### Parabola
The equation of the parabola is given by:
[tex]\[ y = ax^2 + q \][/tex]
The parabola has a turning point at [tex]\((0, 3)\)[/tex]. This tells us that when [tex]\( x = 0 \)[/tex]:
[tex]\[ y = q \][/tex]
So, [tex]\( q = 3 \)[/tex].
The parabola also passes through the point [tex]\((3, 12)\)[/tex].
1. Substitute the point [tex]\((3, 12)\)[/tex] into the equation:
[tex]\[ 12 = a(3)^2 + 3 \][/tex]
Simplifying, we get:
[tex]\[ 12 = 9a + 3 \][/tex]
Subtract 3 from both sides:
[tex]\[ 9 = 9a \][/tex]
[tex]\[ a = 1 \][/tex]
Therefore, the values are:
[tex]\[ a = 1 \][/tex]
[tex]\[ q = 3 \][/tex]
### Summary
For the hyperbola:
[tex]\[ k = 3 \][/tex]
[tex]\[ q = 0 \][/tex]
For the parabola:
[tex]\[ a = 1 \][/tex]
[tex]\[ q = 3 \][/tex]
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