Q. 14 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 7–18, match each graph to one of the following functions:

$$\begin{gathered} \mathit{A}\mathbf{.} y = x^2+ 2 \mathit{B}\mathbf{.} y = -x^2 + 2 \mathit{C}. y = \left|x\right| + 2 \\ \mathit{D}\mathbf{.} y = -\left|x\right|+ 2 \mathit{E}\mathbf{.} y = {(x - 2)}^2 \mathit{F}\mathbf{.} y = -{(x + 2)}^2 \\ \mathit{G}\mathbf{.} y = \left|x-2\right| \mathit{H}\mathbf{.} y = -\left|x+2\right| \mathit{I}\mathbf{.} y = 2x^2 \\ \mathit{J}\mathbf{.} y = -2x^2 \mathit{K}\mathbf{.} y = 2\left|x\right| \mathit{L}\mathbf{.} y = -2\left|x\right| \end{gathered}$$

Step-by-Step Solution

Answer

The given graph is matched to the function $\mathit{C}\mathbf{.}\mathbf{ }y=\left|x\right|+2.$

Step 1. Given Information

There is the given graph we have to match each graph to its functions.

Step 2. Identifying the function of the graph

The given graph is of the function $y=\left|x\right|$ but it is shifted vertically up by $2$ units.

If a positive real number k is added to the outputs 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.

Thus, the function of the graph is $y=\left|x\right|+2$ that matched to option C.