To create the function \( g(x) \) by horizontally stretching the parent function \( f(x) = -12x^2 + 21x - 3 \) with a \( k \) factor of 12, we can use the transformation rule for horizontal stretches. A horizontal stretch by a factor of \( k \) means that we replace \( x \) with \( \frac{x}{k} \). In this case, since \( k = 12 \), we replace \( x \) with \( \frac{x}{12} \) in \( f(x) \). Thus:
\[ g(x) = f\left(\frac{x}{12}\right) \]
Now, let's calculate \( g(x) \) for \( x = 0 \), \( x = 1 \), and \( x = 2 \).
Calculate g(0)
\[ g(0) = f\left(\frac{0}{12}\right) = f(0) = -12(0)^2 + 21(0) - 3 = -3 \]
Calculate g(1)
\[ g(1) = f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3 \] Calculating each term: \[ = -12 \cdot \frac{1}{144} + 21 \cdot \frac{1}{12} - 3 \] \[ = -\frac{12}{144} + \frac{21}{12} - 3 \] \[ = -\frac{1}{12} + \frac{21}{12} - \frac{36}{12} \] \[ = \frac{-1 + 21 - 36}{12} = \frac{-16}{12} = -\frac{4}{3} \]
Calculate g(2)
\[ g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3 \] Calculating each term: \[ = -12 \cdot \frac{1}{36} + 21 \cdot \frac{1}{6} - 3 \] \[ = -\frac{12}{36} + \frac{21}{6} - 3 \] \[ = -\frac{1}{3} + \frac{21}{6} - \frac{18}{6} \] \[ = -\frac{1}{3} + \frac{3}{6} \] \[ = -\frac{1}{3} + \frac{1}{2} = -\frac{2}{6} + \frac{3}{6} = \frac{1}{6} \]
Completed Table 2
Now we will fill in Table 2 with the values calculated:
\[ \begin{array}{|c|c|} \hline x & g(x) \ \hline 0 & -3 \ 1 & -\frac{4}{3} \ 2 & \frac{1}{6} \ \hline \end{array} \]
So the completed Table 2 is:
\[ \begin{array}{|c|c|} \hline x & g(x) \ \hline 0 & -3 \ 1 & -\frac{4}{3} \ 2 & \frac{1}{6} \ \hline \end{array} \]
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