An irrational function has its variable inside a square root, and it arises naturally from finding square roots, as when a square of area S has side √S.
The most basic one, y = √x, is the inverse of y = x² (x ≥ 0), so the two graphs are symmetric about the line y = x; since the expression under the root cannot be negative, the domain and range are each limited to one side.
The graph of y = √(ax + b) + c starts from the point (−b/a, c), where the expression under the root is 0, and when you solve an irrational equation by squaring both sides, roots that do not fit the original can slip in, so you must check every answer.
In the first picture you can overlay y = √x and y = x² to see the symmetry; in the second, the starting point and direction change with a, b and c.
Where do irrational functions come from?
📐 Finding the side of a square
①A square of area S has side √S
②Areas 1, 4 and 9 give neat sides 1, 2 and 3
③But area 2 gives a side of √2 = 1.414…, an irrational number
④A function with its variable inside a square root like this is called an irrational function
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The graph of y = √x and of its inverse y = x² (x ≥ 0), with pairs of points symmetric about the line y = x joined up
🪞 The inverse relationship
①Swap x and y in y = x² (x ≥ 0) to get x = y²
②Taking only y ≥ 0 gives y = √x
③So y = √x is the inverse of the part of y = x² with x ≥ 0
④Inverse functions are symmetric about y = x, and so are these two graphs
⑤In the picture, points with swapped coordinates, such as (4, 2) and (2, 4), form pairs
Properties of y = √x
The basic irrational function
y = √x
Domain x ≥ 0, range y ≥ 0, starts at the origin
Properties of y = √x
Features of the basic form
Domain
The expression under the root must be 0 or more
x ≥ 0
Range
A square root is never negative
y ≥ 0
Increasing or decreasing
As x grows, y grows too
Always increasing
Rate of growth
To raise y by 1, x must grow by 3 (from 1 to 4), then by 5 (from 4 to 9)
Slows down
The graph of y = √(ax + b) + c
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As a, b and c change in y = √(ax + b) + c, the starting point, the direction of the graph, and its domain and range change
1
0
0
General form
y = √(ax + b) + c
From the starting point (−b/a, c) it runs to the right if a > 0 and to the left if a < 0
🔑 Finding the starting point quickly
①Find the x that makes the expression under the root 0: ax + b = 0 gives x = −b/a
②There y = √0 + c = c
③So the starting point is (−b/a, c)
④If a > 0 the domain is x ≥ −b/a and the graph runs right; if a < 0 it is x ≤ −b/a and the graph runs left
⑤Either way the range is y ≥ c
Where an irrational function meets a line
Finding intersections
√(ax + b) + c = mx + n → square both sides and solve
Always put each root you get back into the original equation
⚠️ The trap in squaring both sides
①To solve √x = x − 2, square both sides
②x = (x − 2)² = x² − 4x + 4, so x² − 5x + 4 = 0
③This gives x = 1 or x = 4
④At x = 1 the left side is √1 = 1 but the right side is 1 − 2 = −1, so it does not fit; such a root is called an extraneous root
⑤At x = 4, √4 = 2 and 4 − 2 = 2, so the only solution is x = 4
⑥Squaring turns −1 and 1 into the same 1 and loses the sign, which is why you must check in the original equation
Solving irrational equations
Steps for solving an irrational equation
Step
What to do
Watch out for
Step 1
Get the square-root part on its own
in the form √(…) = (expression)
Step 2
Square both sides
note that the expression must be ≥ 0
Step 3
Simplify and solve
usually a quadratic equation
Step 4
Check in the original equation
discard roots that do not fit (extraneous roots)
Worked examples and unit test
Example 1
Find the domain of the irrational function y = √(2x − 6) + 1.
1
The expression under the root must be at least 0.
2x − 6 ≥ 0
2
Solve the inequality.
x ≥ 3
▸ {x | x ≥ 3}
This graph starts at (3, 1) and runs to the right, and its range is {y | y ≥ 1}.
Example 2
Describe the translation that takes the graph of y = √x to the graph of y = √(x − 2) + 3, and find the starting point.
1
x − 2 in place of x means a shift of 2 in the x-direction, and adding 3 to the whole expression means a shift of 3 in the y-direction.
y = √x → y = √(x − 2) + 3
2
The starting point is (0, 0) moved by the same amounts.
(0, 0) → (2, 3)
▸ A translation of 2 in the x-direction and 3 in the y-direction; the starting point is (2, 3)
Set a = 1, b = −2, c = 3 in the second picture and you get the same graph.
Summary
Basic form
y = √x
Domain x ≥ 0, range y ≥ 0
General form
y = √(ax + b) + c
Starting point (−b/a, c)
Grade-10 school exam type
Find the x-coordinate where y = √(x + 1) meets the line y = x − 1.
①1
②2
③3
④4
⑤5
▸ ③ 3
1
Set the two equal and square both sides (with x − 1 ≥ 0).
√(x + 1) = x − 1 ⇒ x + 1 = (x − 1)²
2
This gives x² − 3x = 0, so x = 0 or 3; the condition x ≥ 1 leaves x = 3.
x² − 3x = 0 ⇒ x = 3 (x = 0 is extraneous)
🎯 Points that often come up
①Set the domain from the condition that the expression under the root is at least 0
②The starting point is (−b/a, c); the graph runs right if a > 0 and left if a < 0
③y = √x is the inverse of y = x² (x ≥ 0), so the graphs are symmetric about y = x
④Solve irrational equations by squaring, then put the roots back into the original equation to weed out extraneous ones
⑤The graph starts at the starting point and grows more and more gently