## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(2, 2)$
In order to find the answer, we will have to recall the following about the graph of $y=f(x)$. a) The graph of the function $y=-f(x)$ involves a reflection about the $x$-axis of the original function $f(x)$. b) The graph of the function $y=f(-x)$ involves a reflection about the $y$-axis of the original function $f(x)$. c) The graph of the function $y=f(x)+a$ defines a vertical shift of $|a|$ units upward when $a \gt 0$, and downward side when $a\lt 0$ of the original function $f(x)$. d) The graph of $y=f(x-p)$ defines a horizontal shift of $|p|$ units to the right when $p \gt 0$, and to the left when $p \lt 0$ of the original function $f(x)$. e) The graph of $y=k\cdot f(x)$ can be obtained a vertical stretch when $k\gt 1$ or compression when $0\lt k \lt1$) of the original function $f(x)$. f) The graph of $y=f(ax)$ can be obtained a horizontal compression when $k\gt 1$ or stretch when $0\lt k \lt1$) of the original function $f(x)$. We will consider point $(f)$ that the resulting function involves a horizontal compression by a factor of $2$ units of the original function $f(x)$. Multiply each $x$-coordinate of $y=f(x)$ by $\frac{1}{a}$ to obtain the same $y$-values. Thus, if $y= f(x)$ contains $(4, 2)$. then $y=f(2x)$ must contain the point $(4\cdot \frac{1}{2}, 2)\longrightarrow (2, 2)$.. Therefore, the answer is not among the choices.