Math 12 Pre-calculus
Irrational functions of the form
x - the inverse of the quadratic
x has the following properties:
The function has a restricted domain: x  0
It has a restricted range: y  0 (by definition, only the positive root is used)
The most basic irrational function y =






It has a graph that is the upper-half of a horizontal parabola
It is the inverse of the right-half of the vertical parabola y = x2
The function approaches infinity, but at a rate that is less than a linear function
(i.e. the curve is always increasing, but is concave down)
Its slope is undefined (vertical slope) at the location (0,0)
In general, irrational functions with linear radicands can be understood by determining these key features:
 The domain – this is critical! Since we can’t root a
 And if we want to graph the function:
negative number, the domain is the solution set to
o The axis intercepts
o The “starting point” – the location on the
radicand  0 .
function where the domain begins. i.e. if
 The range
the domain is x  5 , then the “starting
point” is the location (5,f(5))
Alternatively, we can find the inverse of each irrational function (a quadratic) and use our tools with those functions, instead
1. a) Sketch a graph of the function y 
x  2 by establishing
its domain, range, starting point, and axis intercepts
b) On the same grid, sketch the function y = x – 10.
c) Estimate the solution to
How do we solve an irrational equation? The standard technique is
can be remembered as: “Isolate the radical, then square both sides”
x  3  x 1
x  10  x  2
d) How do we algebraically determine the solution to the equation
in part c)? Read the box on the side.
e) Solve the equation for part c), leaving your answers in exact
form. Did you get two solutions? Are they both valid?
2. Read this: Why did we get an invalid solution to 1e)? By
squaring both sides, we introduced an extraneous solution. Here’s
a simpler example:
Is the radical isolated?
Yep, so square both sides
x  6x  9  x 1
x2  7 x  8  0
2
x
7  17
2
Now manipulate, set to zero
Solve, however you need to. In this
case, quadratic formula
Notice, there are two solutions. You
will need to check them. See Q2
x2  0
 x  2 ( x  2)  0( x  2)
Consider this equation with only one solution, x = 2:
If we multiply both sides by a x+2:
x2  4  0
Expand
Notice, there are two solutions now…
x  2
But the new solution x = -2 is now an extraneous (not valid) solution
Solving irrational equations by squaring both sides can introduce extraneous solutions, so you need to check them all!
3. Solve each equation, making sure to check the validity of your solutions:
a)
x 1  x  7
b)
x  4 x  3
c)
b)
f  x   2x  4  3
c)
x7  x 3
4. Sketch each
a)
f  x   x  7
5. Consider the quadratic function in general form
f  x   x  3  2
f ( x)  x 2  6 x  2 . It has an inverse, but the inverse is not obvious….
a) complete the square on f(x)
b) restrict the domain of f(x) such this it is only half a parabola.
c) determine its inverse function, expressing it in the form y = ….
6. Determine the inverse to each, making sure to state any restrictions such that
they are true inverses of each other:
a)
y  2x2  6x  7
b)
x  y2  2 y  4
c)
y  3x 2  x  11
7. a) Find the equation to the lower half of this parabola (at right). Assume each
square is one unit, vertex is (4.-3), root is 5.
b) Find the inverse of the lower half of this parabola. Be sure to state any required
restrictions.
8. Sketch the function
y  x 2  9 oooh that’s a goodie….