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

Answer 1

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 \( x = 0, 1, 2 \):

1. **For \( x = 0 \)**:
\[
g(0) = f\left(\frac{0}{12}\right) = f(0) = -3.
\]

2. **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:
\[
-12\left(\frac{1}{144}\right) + \frac{21}{12} - 3 = -\frac{1}{12} + \frac{21}{12} - \frac{36}{12} = \frac{-1 + 21 - 36}{12} = \frac{-16}{12} = -\frac{4}{3}.
\]

3. **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:

- \( g(0) = -3 \)
- \( g(1) = -\frac{4}{3} \)
- \( g(2) = \frac{1}{6} \)