Question

The parent function y=x2
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 blank
0 0
2 1
4 blank
answers
16
1
0
4
2

Answer 1

To determine the values in the table for \( y = f(kx) \) where \( f(x) = x^2 \) and \( k = \frac{1}{2} \), we first substitute \( kx \) into the function. The transformation gives us:

\[
y = f(kx) = f\left(\frac{1}{2}x\right) = \left(\frac{1}{2}x\right)^2 = \frac{1}{4}x^2
\]

Now we can calculate \( y \) for the given \( x \) values:

1. For \( x = -4 \):
\[
y = \frac{1}{4}(-4)^2 = \frac{1}{4} \cdot 16 = 4
\]
(This is already provided in the table.)

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

3. For \( x = 0 \):
\[
y = \frac{1}{4}(0)^2 = 0
\]
(This is also provided in the table.)

4. For \( x = 2 \):
\[
y = \frac{1}{4}(2)^2 = \frac{1}{4} \cdot 4 = 1
\]
(This is already provided in the table.)

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

Now we summarize the calculated values for the blanks in the table:

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

Thus, the completed table is:

- For \( x = -2 \), \( y = 1 \).
- For \( x = 4 \), \( y = 4 \).

Final answers for the blanks:

- \( -2 → 1 \)
- \( 4 → 4 \)