Question

The parent function y=x^2 is being horizontally stretched by a k factor of 1/2. Complete the table for y=f(kx)
.(2 points)
x y=f(kx)
-4 4
-2 ___
0 0
2 1
4 ___

Answer 1

To complete the table for the function \( y = f(kx) \) where \( f(x) = x^2 \) and \( k = \frac{1}{2} \), we first need to understand how the transformation affects the function.

When we stretch the function horizontally by a factor of \( k \), we replace \( x \) in our parent function with \( kx \). So, we have:

\[
y = f(kx) = f\left(\frac{1}{2}x\right) = \left(\frac{1}{2}x\right)^2
\]

This simplifies to:

\[
y = \frac{1}{4}x^2
\]

Now we can use this formula to calculate the missing values in the table for given \( x \) values.

1. For \( x = -4 \):
\[
y = \frac{1}{4}(-4)^2 = \frac{1}{4}(16) = 4
\]
(This matches the given value.)

2. For \( x = -2 \):
\[
y = \frac{1}{4}(-2)^2 = \frac{1}{4}(4) = 1
\]

3. For \( x = 0 \):
\[
y = \frac{1}{4}(0)^2 = 0
\]
(This value is already given.)

4. For \( x = 2 \):
\[
y = \frac{1}{4}(2)^2 = \frac{1}{4}(4) = 1
\]
(This matches the given value.)

5. For \( x = 4 \):
\[
y = \frac{1}{4}(4)^2 = \frac{1}{4}(16) = 4
\]

Now, we can fill in the missing values in the table:

\[
\begin{array}{|c|c|}
\hline
x & y = f(kx) \\
\hline
-4 & 4 \\
-2 & 1 \\
0 & 0 \\
2 & 1 \\
4 & 4 \\
\hline
\end{array}
\]

So the completed table is:

\[
\begin{array}{|c|c|}
\hline
x & y = f(kx) \\
\hline
-4 & 4 \\
-2 & 1 \\
0 & 0 \\
2 & 1 \\
4 & 4 \\
\hline
\end{array}
\]

Thus, the values for \( x = -2 \) and \( x = 4 \) are \( 1 \) and \( 4 \), respectively.