Transformations can be applied to functions to change the appearance of
the (slope and intercept) of the function.
The result of the transformation are presented as follows;
- [tex]\begin{tabular}{|c|c|c|c|}f(x-5)&f(x) - 5&-5 \cdot f(x) &f(-5\cdot x)\\3\cdot (x - 5) + 2&3 \cdot x - 5 + 2&-5\cdot (3 \cdot x + 2)&3 \cdot (-5\cdot x) + 2 \end{array}\right][/tex]
Reasons:
The given function is; f(x) = 3·x + 2
The function -5·(3·x + 2) is the same as -5 × f(x) = -f(x)
Therefore;
-5·(3·x + 2) → -5·f(x)
The function 3·x - 5 + 2 = 3·x + 2 - 5 = f(x) - 5
Therefore;
3·x - 5 + 2 → f(x) - 5
The function 3·(x - 5) + 2 by comparison to 3·x + 2 is obtained when x is replaced by (x - 5), therefore;
f(x) = 3·x + 2
f(x - 5) = 3·(x - 5) + 2
3·(x - 5) + 2 → f(x - 5)
The function 3·(-5·x) + 2 is obtained when x in f(x) is replaced by (-5·x),
which gives;
f(x) = 3·x + 2
∴ f(-5·x) = 3·(-5·x) + 2
Which gives;
3·(-5·x) + 2 → f(-5·x)
The completed table is therefore;
[tex]\begin{tabular}{|c|c|c|c|}f(x-5)&f(x) - 5&-5 \cdot f(x) &f(-5\cdot x)\\3\cdot (x - 5) + 2&3 \cdot x - 5 + 2&-5\cdot (3 \cdot x + 2)&3 \cdot (-5\cdot x) + 2 \end{array}\right][/tex]
Learn more about transformation of functions here:
https://brainly.com/question/18076552
