Answer
The graphs are in the attached file.
Work Step by Step
(a) $y=f(x)-1$
If there exist a parent function f(x) and a positive real number ‘c’ such that there is a vertical shift ‘c units’ downward in the graph of the parent function, it follows that:
$y = f(x) - c$
Therefore, in the given question, we have:
A parent function and a related function to the parent function, such that
$y = f(x) - 1$
This implies a vertical shift 1 unit downward.
(b) $y=f(x-1)$
If there exist a parent function f(x) and a positive real number ‘c’ such that there is a horizontal shift ‘c units’ to the right in the graph of the parent function, it follows that:
$y = f(x -c)$
Therefore, in the given question, we have:
A parent function and a related function to the parent function, such that
$y = f(x-1)$
This implies a horizontal shift 1 unit to the right.
(c)$y=\frac{1}{2}f(x)$
If there exist a parent function f(x) and a positive real number ‘c’ such that there is a vertical distortion, the transformation is given as:
$y = cf(x)$
There is a vertical stretch if c > 1 and a vertical shrink if 0 < c < 1. Therefore, in the given question, we have:
A parent function and a related function to the parent function, such that
$y=\frac{1}{2}f(x)$
This implies a vertical shrink of $\frac{1}{2}$ units.
(d) $y=f(-\frac{1}{2}x)$
If there exist a parent function f(x) and a positive real number ‘c’ such that there is a horizontal distortion and a reflection, the transformation is given as:
$y = f(-cx)$
There is a horizontal reflection and shrink if c > 1 and a horizontal stretch if 0 < c < 1 using y-axis as a mirror. Therefore, in the given question, we have:
A parent function and a related function to the parent function, such that
$y=f(-\frac{1}{2}x)$
This implies a horizontal reflection and stretch of $\frac{1}{2}$ units.
