Use the tables to answer the question.

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


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Answer 1

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 use the transformation rule for horizontal stretches. A horizontal stretch by a factor of $k$ means that we replace $x$ with $\frac{x}{k}$. In this case, since $k=12$, we replace $x$ with $\frac{x}{12}$ in $f(x)$. Thus:

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

Now, let's calculate $g(x)$ for $x=0$, $x=1$, and $x=2$.

Calculate g(0)

$$g(0)=f\left(\frac{0}{12}\right)=f(0)=-12(0{)}^2+21(0)-3=-3$$

Calculate g(1)

$$g(1)=f\left(\frac{1}{12}\right)=-12{\left(\frac{1}{12}\right)}^2+21\left(\frac{1}{12}\right)-3$$ Calculating each term: $$=-12\cdot \frac{1}{144}+21\cdot \frac{1}{12}-3$$ $$=-\frac{12}{144}+\frac{21}{12}-3$$ $$=-\frac{1}{12}+\frac{21}{12}-\frac{36}{12}$$ $$=\frac{-1+21-36}{12}=\frac{-16}{12}=-\frac{4}{3}$$

Calculate g(2)

$$g(2)=f\left(\frac{2}{12}\right)=f\left(\frac{1}{6}\right)=-12{\left(\frac{1}{6}\right)}^2+21\left(\frac{1}{6}\right)-3$$ Calculating each term: $$=-12\cdot \frac{1}{36}+21\cdot \frac{1}{6}-3$$ $$=-\frac{12}{36}+\frac{21}{6}-3$$ $$=-\frac{1}{3}+\frac{21}{6}-\frac{18}{6}$$ $$=-\frac{1}{3}+\frac{3}{6}$$ $$=-\frac{1}{3}+\frac{1}{2}=-\frac{2}{6}+\frac{3}{6}=\frac{1}{6}$$

Completed Table 2

Now we will fill in Table 2 with the values calculated:

$$\text{Misplaced hline}$$

So the completed Table 2 is:

$$\text{Misplaced hline}$$