Chapter 2.5, Problem 6AYU

To determine

Whether the graph of $f\left(x\right)=x^3+5$ is obtained by avertical shift of graph of $y=x^3$ to the upward at a distance 5 units.

Answer to Problem 6AYU

The statement is true.

Explanation of Solution

For $c>0$ .

  $h\left(x\right)=f\left(x\right)+c$ shifts the graph of $f\left(x\right)$ vertically upward by $c$ units,

  $h\left(x\right)=f\left(x\right)–c$ shifts the graph of $f\left(x\right)$ vertically downward by $c$ units,

  $h\left(x\right)=f\left(x+c\right)$ shifts the graph of $f\left(x\right)$ horizontally left by $c$ units,

  $h\left(x\right)=f\left(x–c\right)$ shifts the graph of $f\left(x\right)$ horizontally right by $c$ units.

Now, consider the graph of $f\left(x\right)=x^3+5$ .

Here, the graph is of the form $h\left(x\right)=f\left(x\right)+c$ ,

So, $c=5$

The graph of $f\left(x\right)=x^3+5$ can be obtained by a vertical shift of graph of $y=x^3$ to the upward at a distance 5 units.

Hence, the statement is true.