Question

given f(x) and g(x)=f(kx), use the graph to determine the value of k. f(x) g(x) a 5 b $\frac{1}{5}$ c $-\frac{1}{5}$ d - 5

Explanation:

Step1: Recall horizontal - stretch/shrink rule

For a function $y = f(x)$ and $y=f(kx)$, if $|k|> 1$, the graph of $y = f(x)$ is horizontally shrunk by a factor of $\frac{1}{|k|}$, and if $0<|k|<1$, the graph of $y = f(x)$ is horizontally stretched by a factor of $\frac{1}{|k|}$. We can use a point - by - point comparison. Let's assume a point $(x_1,y_1)$ on $f(x)$ and the corresponding point $(x_2,y_2)$ on $g(x)$ such that $y_1 = y_2$ and $y_1=f(x_1)$, $y_2 = f(kx_2)$.

Step2: Select a point on the graphs

Let's consider the y - intercept. For $f(x)$, when $x = 0$, assume $f(0)=y_0$. For $g(x)$, to get the same $y$ - value, if we assume a point on $g(x)$ with the same $y$ - value as the y - intercept of $f(x)$. Let's say a non - zero $x$ value on $g(x)$ corresponding to the y - intercept of $f(x)$. If we look at the horizontal transformation, we can see that the graph of $g(x)$ is a horizontal shrink of $f(x)$. Suppose a point $(x_0, y)$ on $f(x)$ and a point $(x_1,y)$ on $g(x)$ such that $x_0=kx_1$.
Let's assume we can find a point on $f(x)$ say $(5,y)$ and the corresponding point on $g(x)$ with the same $y$ - value is $(1,y)$. Then, since $g(x)=f(kx)$, when $x = 1$ for $g(x)$ and $x = 5$ for $f(x)$ to have the same $y$ - value, we substitute into $g(x)=f(kx)$. If we put $x = 1$ in $g(x)$ and $x = 5$ in $f(x)$ (for the same $y$ - value), we have $5=k\times1$. So $k = 5$.

Answer:

A. 5