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

then what's the answer

Answer 2

To find the function \( g(x) \), we need to apply a horizontal stretch to the parent function \( f(x) = -12x^2 + 21x - 3 \). A horizontal stretch by a factor of \( k \) transforms \( f(x) \) into \( g(x) = f\left(\frac{x}{k}\right) \).

In this case, \( k = 12 \). Therefore, the function \( g(x) \) can be defined as:

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

Now, we will compute \( g(0) \), \( g(1) \), and \( g(2) \) using \( f(x) \):

1. **Calculate \( g(0) \)**:
\[
g(0) = f\left(\frac{0}{12}\right) = f(0)
\]
\[
f(0) = -12(0)^2 + 21(0) - 3 = -3
\]
Thus, \( g(0) = -3 \).

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

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

Putting these results into the second table:

**Table 2**

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