For $f(x) = \sqrt{x}$, how can you manipulate this function's graph to yield this function's graph: $\sqrt{2x+1}$?

Question

The graph of $f(x)=\sqrt{x}$ is the original function. Through transformation, the $f(x)$ function is changed to become $\sqrt{2x+1}$. Why can we not do this:

1. $f(x) = \sqrt{x}$ $\implies$ Graph is straightforward
2. $g(x) = f(x+1) = \sqrt{x+1}$ $\implies$ Graph is $\sqrt{x}$ shifted 1 to the left
3. $k(x) = g(2x) = \sqrt{2x+1}$ $\implies$ 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? How could you then manipulate the graph of $\sqrt{x}$ to give the proper graph?

Answer 1

Your steps 2 and 3 are mixed up! The correct process is:

1. Sketch $f(x)= \sqrt{x}.$

2. Sketch $g(x)=f(2x)=\sqrt{2x}.$

3. Sketch $h(x)=g(x+1/2)=\sqrt{2x+1}.$

You will see that this method gives you the desired starting point of $\left(-\frac{1}{2},0\right)$, since step 2 compresses by a factor of 2, and step 3 shifts the graph to the left by half a unit.