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 in table 2 g(x)
Table 1: x:0, 1, 2 f(x):-3, 6, -9 Table 2: x: 0, 1, 2 g(x):

Answer 1

To create the function \( g(x) \) by horizontally stretching the parent function \( f(x) = -12x^2 + 21x - 3 \) with a stretch factor \( k = 12 \), we replace \( x \) in \( f(x) \) with \( \frac{x}{k} \).

So the function \( g(x) \) is given by:

\[
g(x) = f\left(\frac{x}{12}\right)
\]

Now, we will calculate \( g(x) \) for \( x = 0, 1, 2 \).

First, we substitute \( x = 0 \):

\[
g(0) = f\left(\frac{0}{12}\right) = f(0)
\]

Now, calculate \( f(0) \):

\[
f(0) = -12(0)^2 + 21(0) - 3 = -3
\]

Next, substitute \( x = 1 \):

\[
g(1) = f\left(\frac{1}{12}\right)
\]

Now, calculate \( f\left(\frac{1}{12}\right) \):

\[
f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -12 \cdot \frac{1}{144} + 21 \cdot \frac{1}{12} - 3
\]
\[
= -\frac{12}{144} + \frac{21}{12} - 3
\]
\[
= -\frac{1}{12} + \frac{21}{12} - 3
\]
\[
= \frac{-1 + 21 - 36}{12} = \frac{-16}{12} = -\frac{4}{3}
\]

Next, substitute \( x = 2 \):

\[
g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right)
\]

Now, calculate \( f\left(\frac{1}{6}\right) \):

\[
f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -12 \cdot \frac{1}{36} + 21 \cdot \frac{1}{6} - 3
\]
\[
= -\frac{12}{36} + \frac{21}{6} - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - 3
\]
\[
= -\frac{1}{3} + \frac{21 - 18}{6} = -\frac{1}{3} + \frac{3}{6}
\]
\[
= -\frac{1}{3} + \frac{1}{2} = \frac{-2 + 3}{6} = \frac{1}{6}
\]

Now we can summarize the results in Table 2:

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

Thus, the completed Table 2 is:

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