To find the function \( g(x) \) that is a horizontally stretched version of the parent function \( f(x) = -12x^2 + 21x - 3 \) with a stretch factor \( k = 12 \), we use the concept of a horizontal stretch.
When a function is stretched horizontally by a factor of \( k \), the transformation is given by the function: \[ g(x) = f\left(\frac{x}{k}\right) = f\left(\frac{x}{12}\right) \]
We will calculate \( g(x) \) for the values from table 1.
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For \( x = 0 \): \[ g(0) = f\left(\frac{0}{12}\right) = f(0) = -12(0^2) + 21(0) - 3 = -3 \]
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For \( x = 1 \): \[ g(1) = f\left(\frac{1}{12}\right) = f\left(\frac{1}{12}\right) \] Let's compute \( f\left(\frac{1}{12}\right) \): \[ f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3 \] \[ = -12\left(\frac{1}{144}\right) + 21\left(\frac{1}{12}\right) - 3 \] \[ = -\frac{1}{12} + \frac{21}{12} - 3 \] \[ = \frac{20}{12} - 3 = \frac{20}{12} - \frac{36}{12} = -\frac{16}{12} = -\frac{4}{3} \]
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For \( x = 2 \): \[ g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right) \] Let's compute \( f\left(\frac{1}{6}\right) \): \[ f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3 \] \[ = -12\left(\frac{1}{36}\right) + 21\left(\frac{1}{6}\right) - 3 \] \[ = -\frac{1}{3} + \frac{21}{6} - 3 \] \[ = -\frac{1}{3} + \frac{21}{6} - \frac{18}{6} \] \[ = -\frac{1}{3} + \frac{3}{6} = -\frac{1}{3} + \frac{1}{2} = \left(-\frac{2}{6} + \frac{3}{6}\right) = \frac{1}{6} \]
Now we can fill in the values for table 2:
| x | g(x) | |---|------------| | 0 | -3 | | 1 | -4/3 | | 2 | 1/6 |
So the completed table for \( g(x) \) is:
| x | g(x) | |---|----------| | 0 | -3 | | 1 | -4/3 | | 2 | 1/6 |