Question

The function g(x)
is created by horizontally stretching the parent function, f(x)=−12x^2+21x−3
, with a k factor of 0.5
. Complete the table to show how the function has changed.

Table 1: x-values: 0, 1, 2 and f(x) values=-3, 6, -9
Table 2: x-values: 0, 1, 2 and g(x) values:

Answer 1

To horizontally stretch the function \( f(x) = -12x^2 + 21x - 3 \) with a factor of \( k = 0.5 \), we need to replace \( x \) in the function with \( \frac{x}{k} \). In this case, since \( k = 0.5 \), we will replace \( x \) with \( \frac{x}{0.5} = 2x \).

So, the function \( g(x) \) will be:

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

Let's calculate \( g(x) \) for the specified x-values (0, 1, and 2):

**Calculating \( g(0) \)**:
\[
g(0) = f(2 \cdot 0) = f(0)
\]
\[
f(0) = -12(0)^2 + 21(0) - 3 = -3
\]
So, \( g(0) = -3 \).

**Calculating \( g(1) \)**:
\[
g(1) = f(2 \cdot 1) = f(2)
\]
We already know from the table that \( f(2) = -9 \).
Thus, \( g(1) = -9 \).

**Calculating \( g(2) \)**:
\[
g(2) = f(2 \cdot 2) = f(4)
\]
Calculating \( f(4) \):
\[
f(4) = -12(4^2) + 21(4) - 3
\]
\[
= -12 \times 16 + 84 - 3
\]
\[
= -192 + 84 - 3
\]
\[
= -192 + 81
\]
\[
= -111
\]
So, \( g(2) = -111 \).

Now that we have calculated the values, let's fill in Table 2:

| x-values | g(x) values |
|----------|-------------|
| 0 | -3 |
| 1 | -9 |
| 2 | -111 |

So the final result for Table 2 is:
- g(0) = -3
- g(1) = -9
- g(2) = -111