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
Table 1: x:0, 1, 2 f(x):-3, 6, -9 Table 2: x: 0, 1, 2 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
Table 1: x:0, 1, 2 f(x):-3, 6, -9 Table 2: x: 0, 1, 2 g(x):
Answers
GPT-4o mini
To create the function \( g(x) \) by horizontally stretching the parent function \( f(x) = -12x^2 + 21x - 3 \) with a \( k \) factor of 12, we can modify the input \( x \) in the original function \( f(x) \):
\[
g(x) = f\left(\frac{x}{k}\right) = f\left(\frac{x}{12}\right)
\]
This means we will substitute \( \frac{x}{12} \) into the function \( f(x) \) to compute \( g(x) \).
Let's calculate the values of \( g(x) \) for \( x = 0, 1, 2 \):
1. For \( x = 0 \):
\[
g(0) = f\left(\frac{0}{12}\right) = f(0)
\]
\[
f(0) = -12(0)^2 + 21(0) - 3 = -3
\]
So, \( g(0) = -3 \).
2. For \( x = 1 \):
\[
g(1) = f\left(\frac{1}{12}\right)
\]
Calculating \( f\left(\frac{1}{12}\right) \):
\[
f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -12\left(\frac{1}{144}\right) + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -\frac{12}{144} + \frac{21}{12} - 3
\]
\[
= -\frac{1}{12} + \frac{21}{12} - \frac{36}{12}
\]
\[
= \frac{21 - 36 - 1}{12} = \frac{-16}{12} = -\frac{4}{3}
\]
So, \( g(1) = -\frac{4}{3} \).
3. For \( x = 2 \):
\[
g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right)
\]
Calculating \( f\left(\frac{1}{6}\right) \):
\[
f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -12\left(\frac{1}{36}\right) + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -\frac{12}{36} + \frac{21}{6} - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - \frac{18}{6}
\]
\[
= -\frac{1}{3} + \frac{3}{6} = -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}
\]
So, \( g(2) = \frac{1}{6} \).
Now we can complete Table 2 with the calculated values of \( g(x) \):
\[
\begin{array}{|c|c|c|c|}
\hline
x & 0 & 1 & 2 \\
\hline
g(x) & -3 & -\frac{4}{3} & \frac{1}{6} \\
\hline
\end{array}
\]
### Table 2
| x | 0 | 1 | 2 |
|-----|-----|----------|----------|
| g(x) | -3 | -4/3 | 1/6 |
\[
g(x) = f\left(\frac{x}{k}\right) = f\left(\frac{x}{12}\right)
\]
This means we will substitute \( \frac{x}{12} \) into the function \( f(x) \) to compute \( g(x) \).
Let's calculate the values of \( g(x) \) for \( x = 0, 1, 2 \):
1. For \( x = 0 \):
\[
g(0) = f\left(\frac{0}{12}\right) = f(0)
\]
\[
f(0) = -12(0)^2 + 21(0) - 3 = -3
\]
So, \( g(0) = -3 \).
2. For \( x = 1 \):
\[
g(1) = f\left(\frac{1}{12}\right)
\]
Calculating \( f\left(\frac{1}{12}\right) \):
\[
f\left(\frac{1}{12}\right) = -12\left(\frac{1}{12}\right)^2 + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -12\left(\frac{1}{144}\right) + 21\left(\frac{1}{12}\right) - 3
\]
\[
= -\frac{12}{144} + \frac{21}{12} - 3
\]
\[
= -\frac{1}{12} + \frac{21}{12} - \frac{36}{12}
\]
\[
= \frac{21 - 36 - 1}{12} = \frac{-16}{12} = -\frac{4}{3}
\]
So, \( g(1) = -\frac{4}{3} \).
3. For \( x = 2 \):
\[
g(2) = f\left(\frac{2}{12}\right) = f\left(\frac{1}{6}\right)
\]
Calculating \( f\left(\frac{1}{6}\right) \):
\[
f\left(\frac{1}{6}\right) = -12\left(\frac{1}{6}\right)^2 + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -12\left(\frac{1}{36}\right) + 21\left(\frac{1}{6}\right) - 3
\]
\[
= -\frac{12}{36} + \frac{21}{6} - 3
\]
\[
= -\frac{1}{3} + \frac{21}{6} - \frac{18}{6}
\]
\[
= -\frac{1}{3} + \frac{3}{6} = -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}
\]
So, \( g(2) = \frac{1}{6} \).
Now we can complete Table 2 with the calculated values of \( g(x) \):
\[
\begin{array}{|c|c|c|c|}
\hline
x & 0 & 1 & 2 \\
\hline
g(x) & -3 & -\frac{4}{3} & \frac{1}{6} \\
\hline
\end{array}
\]
### Table 2
| x | 0 | 1 | 2 |
|-----|-----|----------|----------|
| g(x) | -3 | -4/3 | 1/6 |