Radical Functions: Examples (page
2 of 3)
First I'll find the domain by finding where
the argument is non-negative:
Then I'll find some plot points:
(How did I come up with the x-values that gave me such "nice" y-values?
I started with the end values, and worked backwards. For instance, since 25 is a square,
then 3x – 2 = 25 gives me 3x = 27, or x = 9, as my starting value.)
I'll plot the six points from my T-chart
(above), and then I'll sketch my graph:
First I'll find the domain by finding where the
argument is non-negative:
Next, I'll find some plot points — at least
five — for my graph.
Note that, since the graph "starts"
at x = –1/4, the line should not
go below or to the left of the "beginning" of the line. But the function continues
forever in the other direction, so my graph needes to "continue" to the end of
my drawing on the top right-hand "side" of the picture.
I'll first find the domain:
I have to solve the quadratic inside the square root; it may be easier just to look at
the graph of the quadratic.
In this case, x2 + 1
is always above the x-axis (that is, it is always positive), so x can be anything; there is no restriction on the domain of this particular
Next, I'll find some plot
In this case, setting the argument, x2
+ 1, equal to a perfect square and trying
to solve did not work out usefully. Either the input values or the output values were
going to be messy for this function. So I
just used a calculator to find the approximate y-values, accurate
to three decimal places, for my plot-points. This is more than sufficient for graphing.
Finally, I'll do the graph:
Note that, unlike an absolute value graph,
this graph does not have an "elbow" at the bottom; it is curved. Don't go too quickly
when graphing; take the time to notice details like this.