Graphing Radical Functions
Purplemath
The first two square-root graphs on the previous page were in the customary square-root shape of an arc. The last graph on the previous page, though, looked more like a "V", but with a rounded bottom (rather than the sharp-elbowed bottom expected on an absolute-value graph). While most of the square-root functions you'll be given will graph as arcs, some will not, and these other-shaped graphs will become more common if you proceed to calculus.
However, the basic process is still the same; namely, first find the domain, and then plot enough points that you have a good idea of where the graphed line should go. (And, of course, if you have a graphing calculator, check your work!)
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Graph
In the last example on the previous page, the argument of the square root was, as here, a quadratic.
First, I'll find the domain. To do this algebraically, I'll start by setting the argument of the square root to be equal to zero.
16 – x2 = 0
x2 – 16 = 0
(x – 4)(x + 4) = 0
x = –4, 4
But I need to know where the argument is equal to or greater than zero. What do I do now? If I've studied how to solve quadratic inequalities, I can use that methodology. But suppose I haven't learned about that yet?
These two solutions found above are where the argument of the square root crosses the x-axis; that is, they tell me where 16 – x2 is zero. These solutions, or zeroes, split the number line (that is, the x-axis) into three intervals: (–∞, –4), (–4, 4), and (4, ∞). I need the interval(s) on which the 16 – x2 is above the x-axis.
Because I know that the original argument represents a parabola which is above the x-axis in the middle, I know to pick the middle interval for the domain of this square-root function.
(If you're not sure about this, then do a quick graph of y = 16 – x2, and see where the graph is above and below the horizontal axis. The interval where the y-values are positive is (–4, 4), as shown below.)
So the interval between x = –4 and x = 4 will be the domain of this particular radical function. There will be absolutely nothing to graph before or after this interval.
Now I need to find some plot points in addition to the two zeroes that I already have. Because the values didn't work out nicely for this function, I used a calculator to approximate the y-values for my T-chart:
Finally, I'll do my graph:
If you're thinking that my graph above looks a lot like the top half of a circle, then you're exactly right. As a matter of fact, it's the top half of the circle that is centered at the origin and has a radius value of r = 4. (The negative of this square-root function would have given me the bottom half of the same circle.
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Graph
First, I'll find the domain. Because the argument of the radical is a "plus" quadratic, I know that this argument will be positive where the corresponding parabola is above the x-axis. I expect this to be on either side of the x-intercepts, but not in the middle between the intercepts.
The intercepts are found by setting the argument equal to zero, and solving:
x2 – 4x = 0
x (x – 4) = 0
x = 0, 4
These two zeroes of the argument split the number line into three intervals: (–∞, 0), (0, 4), and (4, ∞). I expect the argument to be positive (that is, above the x-axis) on the first and third intervals, but negative (that is, below the x-axis) in the middle. To confirm, I'll look at the graph of parabola y = x2 – 4x:
As I'd expected, the quadratic argument is positive (that is, its graph is higher than the x-axis) before the first intercept at x = 0 and after the second intercept at x = 4.
The argument of the radical is positive on the two unconnected intervals on either end of the parabola's graph. In other words, the domain of the radical is split into TWO pieces which are not attached to each other.
This means that the graph of the radical function will also be in two pieces: one part on the left, stopping at x = 0, and another part on the right, starting at x = 4. There will be nothing but blank space between these two pieces.
Keeping this domain restriction in mind, I'll carefully find some plot-points, using my calculator to get decimal approximations for the y-values:
Finally, I'll do my graph:
Most of the radical-function graphing you'll be doing will involve square roots. But other roots can be graphed, too. Here's an example of a cube-root function:
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Graph
There are no domain constraints with a cube root (or indeed for any other odd-index root), because you can have a "minus" inside a cube root, so I can graph the cube root of a negative number. So, unlike when I'm working with a square root (or any other even-index root), I don't have to start by finding the domain; the domain is going to be "all x".
So I'll go straight to finding some plot points, using my calculator to find decimal approximations:
I was able to find some x-values that gave me nice y-values by setting the argument of the cube root equal to a perfect cube, such as –8 or 1, and solving for x. But, to get enough plot points to be sure of what the graphed line should be doing, I had to use some messy values, too.
My graph looks like this:
Warning: Radicals graph as curved lines. Don't succumb to the temptation of trying to put a straight line through your plotted points. Instead, use enough plot-points to clearly show the shape of the graph, and then draw the graph complete with its curves. You should expect most radical-function graphs to be more wide than they are tall, but don't assume that this will always be the case. Take your time, and remember that it's hard to go wrong with more plot points!
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