Q. 33 - Precalculus Enhanced with Graphing Utilities

Question

If $(1,3)$ is a point on the graph of $y=f\left(x\right)$, which of the following points must be on the graph of$y=2f\left(x\right)$?

Step-by-Step Solution

Answer

The point $(1,6)$lies on the graph of$y=2f\left(x\right)$so the option c is correct 

Step 1. Given data

The point $\left(1,3\right)$lies on the graph of the function$y=f\left(x\right)$

transformed function is$y=2f\left(x\right)$

Step 2. Explanation of correct option C

Point $(1,3)$lies on $y=f\left(x\right)$

so $f\left(1\right)=3$

$y=2f\left(x\right)$is obtained by stretching $y=f\left(x\right)$ vertically

so y coordinate of$y=2f\left(x\right)$at $x=1$

$$\begin{gathered} y=2f\left(x\right) \\ y=2f\left(1\right) \\ y=2\left(3\right) \\ y=6 \end{gathered}$$

point$(1,6)$ lies on$y=2f\left(x\right)$ so option c is correct.

Step 3. explanation of incorrect options

• y coordinate of$y=2f\left(x\right)$ is two times of $y=f\left(x\right)$so point$\left(1,\frac{3}{2}\right)$can not lie on it
• y coordinate of $y=2f\left(x\right)$ is two times of$y=f\left(x\right)$ so point $(2,3)$can not lie on it
• y coordinate of $y=2f\left(x\right)$ is two times of $y=f\left(x\right)$ so point$\left(\frac{1}{2},3\right)$can not lie on it