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)
this table
( \begin{array}{|c|c|}hline x & y = kf(x) \\ hline 0 & 12 \\ hline 1 & 15 \\ hline 2 & 18 \\ hline end{array} )
this table
( \begin{array}{|c|c|}hline x & y = kf(x) \\ hline 0 & -12 \\ hline 1 & -15 \\ hline 2 & -18 \\ hline end{array} )
this table
( \begin{array}{|c|c|}hline x & y = kf(x) \\ hline 0 & 1.3 \\ hline 1 & 1.7 \\ hline 2 & 2 \\ hline end{array} )
this table
( \begin{array}{|c|c|}hline x & y = kf(x) \\ hline 0 & 4 \\ hline 1 & 5 \\ hline 2 & 6 \\ hline end{array} )

Explanation:

Step1: Find the value of \(y = |x + 4|\) at \(x = 0\)

Substitute \(x = 0\) into \(y=|x + 4|\), we get \(y=|0 + 4|=4\).

Step2: Check the first table

In the first table, when \(x = 0\), \(y = 12\). Since \(y=kf(x)\) and \(f(0)=4\), then \(k=\frac{y}{f(x)}=\frac{12}{4}=3\).

Step3: Verify for \(x = 1\) and \(x = 2\)

For \(x = 1\), \(f(1)=|1 + 4|=5\), and in the first table \(y = 15\), \(k=\frac{15}{5}=3\).
For \(x = 2\), \(f(2)=|2+4| = 6\), and in the first table \(y = 18\), \(k=\frac{18}{6}=3\).

Answer:

The first table (with \(x = 0,y = 12\); \(x = 1,y = 15\); \(x = 2,y = 18\))