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
= 27, or x
= 9, as my starting
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.
first find the domain: I have to solve the quadratic inside the
square root; it may be easier just to look at the
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 square-root function.
I'll find some plot points:
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
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.