Radical Functions
In the previous section, we mentioned radical functions. We are going to look a little more into
them here. Radical functions are of the form:
𝑛
𝑦 = √𝑥 .
The value in the radical can be just x, or it can be an expression such as 3𝑥 2 − 2𝑥 +
3, 𝑠𝑖𝑛𝑥, 3𝑥 − 4, 𝑒𝑡𝑐.
If the index of the root (that’s the little ‘n’) is odd, then the expression underneath the radical
can attain any value: positive, negative or zero. The restrictions placed on the domain are not
augmented by the radical.
If the index of the root is even, then the expression underneath the radical must be
nonnegative. (NOTE: ‘Nonnegative’ means the value is either positive or zero). To find the
domain of these functions, we create an inequality with the expression under the radical and
assert that it is greater than or equal to zero. Then we solve the inequality.
Examples: Find the domain of each function below
5
1. 𝑓(𝑥) = √3𝑥 2 − 5𝑥 + 7 Since the index is 5, this is an odd root. There are no
restrictions placed on the domain by the radical so we just look at the expression
underneath. We have a polynomial underneath, so the domain here is all real numbers.
We can put any value we wish in for x and will get a function value out. In interval
notation, our domain is: (−∞, ∞).
4
2. 𝑓(𝑥) = √4 − 𝑥 2 Here the index is 4. This means we must restrict the domain such that
only nonnegative values occur under the radical.
4 − 𝑥2 ≥ 0
4 ≥ 𝑥2
𝑥2 ≤ 4
|𝑥| ≤ 2
So our domain is the set of x values whose absolute value is less than or equal to 2. In
interval notation, this is: [-2,2].
3. 𝑓(𝑥) = √2𝑥 2 − 5𝑥 − 3 We again have an even root. This time, we can approach
finding the acceptable values in two ways. We can graph the expression under the
radical in a calculator and see where it is nonnegative, or we can try to find its zeros
algebraically and deduce from there where it is nonnegative. We will assume that the
calculator method is self explanatory and look at the algebra.
2𝑥 2 − 5𝑥 − 3 = 0
(2𝑥 + 1)(𝑥 − 3) = 0
(2𝑥 + 1) = 0 𝑜𝑟 (𝑥 − 3) = 0
−1
𝑥=
𝑜𝑟 𝑥 = 3
2
Now that we have these zeros, we know that the polynomial (the expression under the
radical) will only be able to change signs at
−1
2
𝑎𝑛𝑑 3. So we can look between these
numbers to see whether the polynomial is positive or negative. Whatever it is between
the zeros, it will be the opposite outside of them (this is because of the nature of the
second degree polynomial- we will discuss this in a later section of the course). The
number 0 is between
−1
2
𝑎𝑛𝑑 3, so let’s use it.
Putting 0 in, we get: 2(0)2 − 5(0) − 3 = −3. Since the expression is negative here, we
will use the values outside this region, that is: 𝑥 ≤
−1
2
𝑜𝑟 𝑥 ≥ 3, because the values
inside the region will give us nonreal values. In interval notation, this would be:
(−∞,
−1
2
] ∪ [3, ∞).
-4 -3 -2 -1
0
1 2
3 4
Shifting
In addition to looking at the domains of these functions, we must also discuss transformations
𝑛
of the graph of 𝑦 = √𝑥. We will look specifically at just the square root, although the results
are general to all radical functions.
Let’s start with the function 𝑦 = √𝑥. Now consider two numbers h and k. We’ll use h to
translate the graph horizontally, and k to translate vertically.
The graph of 𝑦 = √𝑥 − ℎ + 𝑘 would be shifted horizontally h units and vertically k units. Please
note the signs.
𝑦 = √𝑥 − 6 + 5 would take the original graph and move it to the right 6 units and up 5 units.
𝑦 = √𝑥 + 3 + 2 would take the original graph and move it to the left 3 units and up 2 units.
Stretching and shrinking
We can also stretch and shrink the graph. We do this by multiplying by a constant. If the
multiplication occurs inside the radical, it is a horizontal transformation. If it occurs outside the
radical, it is vertical.
For horizontal transformations using a multiple of ‘a’: if a>1, it is a horizontal shrink
If 0<a<1, it is a horizontal stretch
For vertical transformations using a multiple of ‘b’: if b>1, it is a vertical stretch
If 0<b<1, it is a vertical shrink.
Please note that with nonperiodic functions (like the square root) it is hard to tell the difference
between a vertical shrink and a horizontal stretch or between a vertical stretch and a horizontal
shrink. The equations help more than the graphs.
Look again at the graph of 𝑦 = √𝑥.
If we consider 𝑦 = √5𝑥, it will take the original graph and shrink it horizontally by a factor of 5.
This would mean that the new function will cover values 5 times as fast as the original. When
x=1, the new function will treat it as though it were 5. So we have covered all the function
values between 0 and 1 that the original function covers between 0 and 5.
If we consider 𝑦 = 5√𝑥, it will take the original graph and stretch it vertically by a factor of 5.
This will quintuple the value of each function value. When x=1, the new function will have a
value of 5, instead of the function value of 1 which the original has.
Reflections
The last transformation we consider is reflecting the graph. In order to reflect the graph over
the x axis, we multiply the function by -1. This essentially takes the opposite of all the y values.
To reflect the graph over the y axis, we multiply the x value by -1.
Reflect 𝑦 = √𝑥 over the x axis. All we need do is use 𝑦 = −√𝑥.
To reflect 𝑦 = √𝑥 over the y axis, we would use 𝑦 = √−𝑥.
Examples: For each equation below, describe the transformation that has been done, and give
the domain of the new function.
1. 𝑦 = 3√𝑥 − 5 + 4
This function has been shifted right 5 units, up 4 units and vertically stretched by a
factor of 3. The domain is [5, ∞).
2. 𝑦 = −√2𝑥 − 3
This function has been horizontally shrunken by a factor of 2, reflected in the x axis, and
shifted down 3 units. The domain is [0, ∞).
5
3. 𝑦 = √3𝑥 + 5 = √3(𝑥 + 3)
This function has been horizontally shrunken by a factor of 3 and shifted left 5/3 units.
−5
The domain is [ 3 , ∞).
1
−1
4. 𝑦 = −√− 2 𝑥 + 4 − 2 = −√ 2 (𝑥 − 8) − 2
This function has been reflected in the y axis, horizontally stretched by a factor of 1/2,
shifted right 8 units, reflected in the x axis, and shifted down 2 units. The domain is
(−∞, 8].
Notice that it is easier to factor the terms in the radical to get a better idea of what is going on.