Square Root Functions and Their Graphs

A photographer editing an image adjusts the brightness of each pixel. One common adjustment formula maps an input brightness value $x$ to an output brightness \sqrt. This is done deliberately — square root curves compress bright regions and expand dark ones, making images look more natural to the human eye. The shape of that curve is specific and predictable, and understanding it is the entire point of this lesson.

The parent function is y = \sqrt. Before graphing anything, ask where this function is even defined. You cannot take the square root of a negative number and get a real output, so $x$ must be greater than or equal to zero. That single restriction gives the function its domain: $[0, \infty)$. The range follows immediately — since KaTeX can only parse string typed expression for all valid inputs, the output is never negative, so the range is also $[0, \infty)$.

The graph starts at the origin, rises steeply at first, then flattens out. It never turns around. It has no vertex in the parabola sense, but it does have a definite starting point — the endpoint $(0, 0)$ — and that point matters on the exam.

Now for transformations. The general form of a transformed square root function is

Each parameter does something specific. The value $h$ shifts the graph horizontally — adding to $x$ inside the radical moves the starting point left, subtracting moves it right. The value $k$ shifts the graph vertically — positive $k$ moves it up, negative moves it down. The value $a$ stretches or compresses the graph vertically, and when $a$ is negative, it reflects the graph across the $x$-axis so the curve opens downward instead of upward.

The starting point of the transformed graph is always $(h, k)$. That is the new endpoint, the corner where the graph begins. The domain becomes $x \geq h$, written as $[h, \infty)$. The range depends on $a$: if $a > 0$, the range is $[k, \infty)$; if $a < 0$, the range is $(-\infty, k]$.

Now consider what happens when $a$ is negative. The graph reflects over the $x$-axis and the curve bends downward from its starting point.

One more situation comes up on the exam: identifying the equation from a graph. Look at where the graph starts — that gives you $h$ and $k$. Look at whether it goes up or down — that tells you the sign of $a$. Then substitute one other point from the graph to find the exact value of $a$.