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 need to adjust the input of the function \( f \).

The transformation for a horizontal stretch by a factor of \( k \) is given by:
\[
g(x) = f\left(\frac{x}{k}\right)
\]
In this case, \( k = 12 \), so:
\[
g(x) = f\left(\frac{x}{12}\right)
\]

Now we will compute the values of \( g(x) \) based on the values from the table of \( f(x) \).

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

### Step 2: Determine \( g(1) \)
For \( x = 1 \):
\[
g(1) = f\left(\frac{1}{12}\right)
\]

First, calculate \( f\left(\frac{1}{12}\right) \):
\[
f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -12\left(\frac{1}{144}\right) + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -\frac{1}{12} + \frac{21}{12} - 3
\]
\[
= \frac{20}{12} - 3 = \frac{20}{12} - \frac{36}{12} = -\frac{16}{12} = -\frac{4}{3}
\]
Thus, \( g(1) = -\frac{4}{3} \).

### Step 3: Determine \( g(2) \)
For \( x = 2 \):
\[
g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right)
\]

Next, compute \( f\left(\frac{1}{6}\right) \):
\[
f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -12\left(\frac{1}{36}\right) + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - \frac{18}{6}
\]
\[
= -\frac{1}{3} + \frac{3}{6} = -\frac{1}{3} + \frac{1}{2}
\]
Convert \( -\frac{1}{3} \) to sixths:
\[
= -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}
\]
Thus, \( g(2) = \frac{1}{6} \).

### Final table values for \( g(x) \):
- \( g(0) = -3 \)
- \( g(1) = -\frac{4}{3} \)
- \( g(2) = \frac{1}{6} \)

So, the completed Table 2 looks like this:

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