To create the function \( g(x) \) from the function \( f(x) = -12x^2 + 21x - 3 \) by horizontally stretching it with a factor of \( k = 12 \), we use the formula for horizontal stretching, which modifies the input of the function:
\[ g(x) = f\left(\frac{x}{k}\right) = f\left(\frac{x}{12}\right). \]
We will now compute the values of \( g(x) \) for \( x = 0, 1, 2 \):
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For \( x = 0 \): \[ g(0) = f\left(\frac{0}{12}\right) = f(0) = -3. \]
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For \( x = 1 \): \[ g(1) = f\left(\frac{1}{12}\right). \]
First, calculate \( f\left(\frac{1}{12}\right) \): \[ f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3. \]
Calculating each term: \[ -12\left(\frac{1}{144}\right) + \frac{21}{12} - 3 = -\frac{1}{12} + \frac{21}{12} - \frac{36}{12} = \frac{-1 + 21 - 36}{12} = \frac{-16}{12} = -\frac{4}{3}. \]
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For \( x = 2 \): \[ g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right). \]
Now calculate \( f\left(\frac{1}{6}\right) \): \[ f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3. \]
Calculating each term: \[ -12\left(\frac{1}{36}\right) + \frac{21}{6} - 3 = -\frac{1}{3} + \frac{21}{6} - \frac{18}{6} = -\frac{1}{3} + \frac{3}{6} = -\frac{1}{3} + \frac{1}{2}. \] Converting \(-\frac{1}{3}\) and \(\frac{1}{2}\) to a common denominator: \[ -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}. \]
The completed Table 2 is as follows:
| x | g(x) | |---|------------| | 0 | -3 | | 1 | -\(\frac{4}{3}\) | | 2 | \(\frac{1}{6}\) |
To summarize:
- \( g(0) = -3 \)
- \( g(1) = -\frac{4}{3} \)
- \( g(2) = \frac{1}{6} \)
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