Chapter E, Problem 1E

To determine

To fill: The blank in the statement “If k is a positive real number, to obtain the graph of $y=f\left(x\right)+k$ from the graph of $y=f\left(x\right)$, shift the graph of $y=f\left(x\right)$ ______ k units. To obtain the graph of $y=f\left(x\right)-k$ from the graph of $y=f\left(x\right)$, shift the graph of $y=f\left(x\right)$ ______ k units.”

Answer to Problem 1E

The completed statement is “If k is a positive real number, to obtain the graph of $y=f\left(x\right)+k$ from the graph of $y=f\left(x\right)$, shift the graph of $y=f\left(x\right)$ up k units. To obtain the graph of $y=f\left(x\right)-k$ from the graph of $y=f\left(x\right)$, shift the graph of $y=f\left(x\right)$ down k units”.

Explanation of Solution

Result used:

If a positive real number k is added to the output of a function, $y=f\left(x\right)$, the graph of the new function, $y=f\left(x\right)+k$, is the graph of f shifted vertically up k units.

If a positive real number k is subtracted from the output of a function, $y=f\left(x\right)$, the graph of the new function, $y=f\left(x\right)-k$, is the graph of f shifted vertically down k units.

Description:

Consider the function $y=f\left(x\right)+k$, where k is a positive real number.

Clearly, the number k is added to the output of the function $y=f\left(x\right)$.

By the above result, the graph of $y=f\left(x\right)+k$ is the graph of f shifted vertically up k units.

Similarly, consider the function $y=f\left(x\right)-k$, where k is a positive real number.

Clearly, the number k is subtracted from the output of the function $y=f\left(x\right)$.

By the above result, the graph of $y=f\left(x\right)-k$ is the graph of f shifted vertically down k units.

Therefore, the completed statement is “If k is a positive real number, to obtain the graph of $y=f\left(x\right)+k$ from the graph of $y=f\left(x\right)$, shift the graph of $y=f\left(x\right)$ up k units. To obtain the graph of $y=f\left(x\right)-k$ from the graph of $y=f\left(x\right)$, shift the graph of $y=f\left(x\right)$ down k units”.