Question

Which of the following tables shows that y=kf(x)
results in the parent function y=|x+4|
being vertically stretched by a k factor of 3?(1 point)
Responses

x y=kf(x)
0 −12
1 −15
2 −18
x y equals k f left parenthesis x right parenthesis 0 negative 12 1 negative 15 2 negative 18
table 1
x y=kf(x)
0 4
1 5
2 6
x y equals k f left parenthesis x right parenthesis 0 4 1 5 2 6
table 2
x y=kf(x)
0 1.3
1 1.7
2 2
x y equals k f left parenthesis x right parenthesis 0 1.3 1 1.7 2 2
table3
x y=kf(x)
0 12
1 15
2 18
table 4

Answer 1

To identify which table demonstrates that \( y = kf(x) \) results in the parent function \( y = |x + 4| \) being vertically stretched by a factor of 3, we begin by determining the output values of the parent function at the given \( x \) values, where \( f(x) = |x + 4| \).

Calculating the outputs for \( f(x) \):

- For \( x = 0 \): \( f(0) = |0 + 4| = 4 \)
- For \( x = 1 \): \( f(1) = |1 + 4| = 5 \)
- For \( x = 2 \): \( f(2) = |2 + 4| = 6 \)

Next, if we apply a vertical stretch by a factor of \( k = 3 \), the transformation will change the outputs to:

- For \( x = 0 \): \( y = 3f(0) = 3 \times 4 = 12 \)
- For \( x = 1 \): \( y = 3f(1) = 3 \times 5 = 15 \)
- For \( x = 2 \): \( y = 3f(2) = 3 \times 6 = 18 \)

Now we can compare these transformed output values (12, 15, 18) to the tables provided:

- **Table 1**: \( 0 \to -12, 1 \to -15, 2 \to -18 \) (not what we're looking for)
- **Table 2**: \( 0 \to 4, 1 \to 5, 2 \to 6 \) (this represents the parent function, not the stretched version)
- **Table 3**: \( 0 \to 1.3, 1 \to 1.7, 2 \to 2 \) (not relevant)
- **Table 4**: \( 0 \to 12, 1 \to 15, 2 \to 18 \) (this matches our calculated values)

Thus, **Table 4** shows that \( y = kf(x) \) results in the parent function \( y = |x + 4| \) being vertically stretched by a factor of 3.