Question

Use the tables to answer the question.

The function g(x)
is created by horizontally stretching the parent function, f(x)=−12x2+21x−3
, with a k factor of 12
. Complete the table to show how the function has changed. (1 point)
Table 1

x
f(x)
0
−3
1
6
2
−9

Table 2

x
g(x)
0
1
2


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Answer 1

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:

\[
\begin{array}{|c|c|}
\hline
x & g(x) \\
\hline
0 & -3 \\
1 & -\frac{4}{3} \\
2 & \frac{1}{6} \\
\hline
\end{array}
\]