which of the following tables shows that ( y = kf(x) ) results in the parent function ( y = |x + 4| ) being…

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

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

Answer

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))