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 find the function \( g(x) \) that is a horizontally stretched version of the parent function \( f(x) = -12x^2 + 21x - 3 \) with a stretch factor \( k = 12 \), we use the concept of a horizontal stretch.

When a function is stretched horizontally by a factor of \( k \), the transformation is given by the function:
\[
g(x) = f\left(\frac{x}{k}\right) = f\left(\frac{x}{12}\right)
\]

We will calculate \( g(x) \) for the values from table 1.

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

2. For \( x = 1 \):
\[
g(1) = f\left(\frac{1}{12}\right) = f\left(\frac{1}{12}\right)
\]
Let's compute \( f\left(\frac{1}{12}\right) \):
\[
f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -12\left(\frac{1}{144}\right) + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -\frac{1}{12} + \frac{21}{12} - 3
\]
\[
= \frac{20}{12} - 3 = \frac{20}{12} - \frac{36}{12} = -\frac{16}{12} = -\frac{4}{3}
\]

3. For \( x = 2 \):
\[
g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right)
\]
Let's compute \( f\left(\frac{1}{6}\right) \):
\[
f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -12\left(\frac{1}{36}\right) + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - \frac{18}{6}
\]
\[
= -\frac{1}{3} + \frac{3}{6} = -\frac{1}{3} + \frac{1}{2} = \left(-\frac{2}{6} + \frac{3}{6}\right) = \frac{1}{6}
\]

Now we can fill in the values for table 2:

| x | g(x) |
|---|------------|
| 0 | -3 |
| 1 | -4/3 |
| 2 | 1/6 |

So the completed table for \( g(x) \) is:

| x | g(x) |
|---|----------|
| 0 | -3 |
| 1 | -4/3 |
| 2 | 1/6 |