The graph of $f(x) = \sqrt x$ is the original function. Through transformation, $f(x)$ is changed to become $\sqrt{2x+1}$. Why can we not do the following:
- $f(x) = \sqrt x$ ==> Graph is straighforward
- $g(x) = f(x+1) = \sqrt{x+1}$ ==> Graph is $\sqrt x$ shifted $1$ to the left
- $k(x) = g(2x) = \sqrt{2x+1}$ ==> Graph is $\sqrt{x+1}$ compressed by factor of $2$.
But the proper graph of $\sqrt{2x+1}$ is a graph of $\sqrt x$ starting at point $(\frac{-1}{2}, 0)$. Why does my method not work? But if you scale first, then shift the function:
- $f(x) = \sqrt x$
- $g(x) = f(2x)$
- $k(x) = g(x+\frac{1}{2})$
the proper graph is the result.
First you have to bring forward the factor two:
$$\sqrt{2x+1}=\sqrt{2\left(x+\frac12\right)}=g\left(x+\frac12\right)$$
where
$$g(x)=\sqrt{2x}.$$
$g(x)$ is easy to plot then so is it $g\left(x+\frac12\right).$
In the following figure the blue line is $f(x)=\sqrt{x}$, the purple line is $g(x)=f(2x)$ and the green line is $g\left(x+\frac12\right)=f\left(2(x+\frac12)\right)=f(2x+1)=\sqrt{2x+1}.$