Q. 11 - 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 $\mathit{I}\mathbf{.}\mathbf{ }y=2x^2.$

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 similar to the graph of the function $y=x^2$ but it is stretched vertically by a factor of $2.$

When the right side of a function $y=f\left(x\right)$ is multiplied by a positive number a,

the graph of the new function $y=af\left(x\right)$ is obtained by multiplying each y-coordinate on the graph of $y=f\left(x\right)$ by a. The new graph is a vertically stretched $(if a>1)$ version of the graph of $y=f\left(x\right).$

Thus, the function of the graph is $y=2x^2$that matched to option I.