Grade by Grade
A. Characteristics of Functions · Functions and Equivalent Expressions

Inverses of Quadratic Functions and Restricted Domains

A1.5 A1.6 A1.7About 70 minutes
Learning goals
  • I can find the inverse of a quadratic function algebraically, graphically and numerically.
  • I can explain why the inverse of a quadratic function is not a function, and restrict the domain so that it is.
  • I can relate the domain and range of a quadratic function to those of its inverse.
  • I can use the inverse of a quadratic in a real-world situation.

Words to know

restricted domain
A domain that has been limited (for example to x ≥ 2) so the function has a certain property, such as an inverse that is a function.
± (plus or minus)
Shows both a positive and a negative value, as in √25 = 5 but x² = 25 means x = ±5.
vertex form
y = a(x − h)² + k, with vertex (h, k). The best form for finding a quadratic's inverse.
branch
One of the separate parts of a graph, such as the top half y = √x of the relation y = ±√x.

Undoing a square

If a square floor tile has area 36 cm², how long is a side? You undo squaring by taking a square root: 6 cm. But in pure algebra, x² = 36 has two solutions, 6 and −6. That little ± is the whole story of this lesson. The inverse of a quadratic function is not a function unless we cut the parabola in half.

The inverse of f(x) = x²

Swap x and y: x = y². Solve for y: y = ±√x.

This inverse relation has two branches:

  • the top branch g(x) = √x (range y ≥ 0), and
  • the bottom branch h(x) = −√x (range y ≤ 0).

Together they make the sideways parabola from Lesson 1, which fails the vertical line test. Each branch on its own is a function.

Compare the domains and ranges:

  • f(x) = x²: D = {x ∈ ℝ}, R = {y ∈ ℝ | y ≥ 0}
  • inverse relation y = ±√x: D = {x ∈ ℝ | x ≥ 0}, R = {y ∈ ℝ}

The domain and range have swapped, just as they always do.

Restricting the domain

To make the inverse a function, restrict the domain of the quadratic to one side of its vertex, so that it passes the horizontal line test.

  • f(x) = x² with x ≥ 0 has inverse f⁻¹(x) = √x.
  • f(x) = x² with x ≤ 0 has inverse f⁻¹(x) = −√x.

The vertex is where the parabola turns around, so it is the natural place to cut: keeping everything on one side of it keeps as much of the parabola as possible.

f(x) = (x − 2)² − 5x− 2square− 5yinverse: undo in reverse ordery+ 2± √+ 5xy = 2 ± √(x + 5): the ± gives two outputs, so not a function
To undo f(x) = (x − 2)² − 5, add 5, take the square root (both signs), then add 2.
Worked example

Example 1: A quadratic in vertex form

Find the inverse of f(x) = (x − 2)² − 5. Is it a function? Then restrict the domain of f so that its inverse is a function, and state the domain and range of f and f⁻¹.

  1. Write y = (x − 2)² − 5 and swap: x = (y − 2)² − 5.
  2. Add 5: x + 5 = (y − 2)².
  3. Take the square root of both sides, with ±: ±√(x + 5) = y − 2.
  4. Add 2: y = 2 ± √(x + 5). For x = 4 this gives y = 2 + 3 = 5 and y = 2 − 3 = −1: two outputs, so the inverse is not a function.
  5. The vertex of f is (2, −5). Restrict f to x ≥ 2 (the right half). Its outputs then satisfy y ≥ −5.
  6. For the right half, the inverse uses only the + sign (its outputs must be ≥ 2): f⁻¹(x) = 2 + √(x + 5).
  7. f (restricted): D = {x ∈ ℝ | x ≥ 2}, R = {y ∈ ℝ | y ≥ −5}. f⁻¹: D = {x ∈ ℝ | x ≥ −5}, R = {y ∈ ℝ | y ≥ 2}.

Inverse relation y = 2 ± √(x + 5) (not a function). With x ≥ 2, f⁻¹(x) = 2 + √(x + 5).

xy−4−268−4−268(2, −5)(−5, 2)f, x ≥ 2f⁻¹y = xf(x) = (x − 2)² − 5, x ≥ 2D = {x ∈ ℝ | x ≥ 2}R = {y ∈ ℝ | y ≥ −5}f⁻¹(x) = 2 + √(x + 5)D = {x ∈ ℝ | x ≥ −5}R = {y ∈ ℝ | y ≥ 2}
The right half of f (x ≥ 2, thick curve) reflects in y = x to the top branch y = 2 + √(x + 5). The faint halves are the parts left out by the restriction. The vertex (2, −5) reflects to (−5, 2).

Choosing + or −

After restricting, decide the sign by asking: what values must the inverse output? The inverse outputs the restricted domain of f. If f was restricted to x ≥ 2, the inverse must give values ≥ 2, so use 2 + √(...). If f was restricted to x ≤ 2, use 2 − √(...).

Worked example

Example 2: Standard form first

Find the inverse of f(x) = x² + 6x + 4 and state its domain.

  1. Swapping right away gives x = y² + 6y + 4, which is hard to solve for y. So first complete the square:
  2. x² + 6x + 4 = (x² + 6x + 9) − 9 + 4 = (x + 3)² − 5.
  3. Swap: x = (y + 3)² − 5 → x + 5 = (y + 3)² → y + 3 = ±√(x + 5) → y = −3 ± √(x + 5).
  4. The range of f is y ≥ −5 (vertex (−3, −5), opens up), so the domain of the inverse is {x ∈ ℝ | x ≥ −5}.

Inverse relation y = −3 ± √(x + 5), with domain x ≥ −5.

Worked example

Example 3: When a ≠ 1 or a < 0

Find the inverse of (a) f(x) = 2(x − 1)² + 3 (b) g(x) = −(x + 1)² + 4.

  1. (a) Swap: x = 2(y − 1)² + 3 → x − 3 = 2(y − 1)² → (x − 3)/2 = (y − 1)² → y = 1 ± √((x − 3)/2). Domain of the inverse: x ≥ 3 (the range of f).
  2. (b) Swap: x = −(y + 1)² + 4 → x − 4 = −(y + 1)² → 4 − x = (y + 1)² (multiply both sides by −1) → y = −1 ± √(4 − x).
  3. g opens down with vertex (−1, 4), so its range is y ≤ 4. The inverse needs 4 − x ≥ 0, so its domain is x ≤ 4. The two facts agree.

(a) y = 1 ± √((x − 3)/2), x ≥ 3. (b) y = −1 ± √(4 − x), x ≤ 4.

Worked example

Example 4: A falling acorn

An acorn falls from a branch of a tall oak tree. The distance it has fallen after t seconds is d(t) = 4.9t² metres, for t ≥ 0 (ignoring air resistance). Find the inverse, and use it to find when the acorn has fallen 9.8 m.

  1. Solve for t, keeping the meaning of the letters: t² = d/4.9, so t = ±√(d/4.9).
  2. Time cannot be negative here (the domain of d(t) is t ≥ 0), so keep only the + sign: t(d) = √(d/4.9).
  3. t(9.8) = √(9.8/4.9) = √2 ≈ 1.41 s.

t(d) = √(d/4.9); the acorn has fallen 9.8 m after about 1.41 s. The context restricts the domain, so the inverse is a function.

Just for fun

Riddle: what has two square roots, one vertex and an identity crisis when you reflect it in y = x?

A parabola. It walks in as a function and comes out as a sideways relation.

Hands-on activity

Fold a parabola in half

You need: grid paper, a ruler, a pencil, and tracing paper (or a window to trace against).

  1. Graph f(x) = (x − 2)² − 5 using x = −1, 0, 1, 2, 3, 4, 5. Draw the dashed line y = x.
  2. Swap each point's coordinates and plot the inverse relation. Join the points smoothly.
  3. Check with a vertical ruler: is the inverse a function?
  4. Now colour only the part of f with x ≥ 2 and the matching part of the inverse.
  5. Check: f⁻¹(4) should be 5 and f⁻¹(−1) should be 4. Use the formula f⁻¹(x) = 2 + √(x + 5) to confirm.

What to look for: f gives (−1, 4), (0, −1), (1, −4), (2, −5), (3, −4), (4, −1), (5, 4). The inverse points are (4, −1), (−1, 0), (−4, 1), (−5, 2), (−4, 3), (−1, 4), (4, 5). The vertical line x = 4 hits both (4, −1) and (4, 5), so the full inverse is not a function. The coloured halves both pass the test. Formula check: 2 + √9 = 5 and 2 + √4 = 4. ✓

Your turn

1

Find the inverse of f(x) = x² + 7. Is it a function?

2

Find the inverse of f(x) = (x + 4)².

3

Find the inverse of f(x) = (x − 3)² + 1 and state its domain.

4

How should you restrict f(x) = (x − 3)² + 1 so that its inverse is f⁻¹(x) = 3 + √(x − 1)?

5

State the domain and range of f(x) = x² − 2 and of its inverse relation.

6

Why must you write ± when you take the square root of both sides of an equation?

More practice: warm-up

1

Solve y² = 49.

2

What is the inverse relation of y = x²?

3

Is the inverse of f(x) = x², x ≥ 0, a function? What is it?

4

Where should you cut a parabola to make its inverse a function?

5

The range of a quadratic is y ≥ 6. What is the domain of its inverse?

More practice: on your own

1

Find the inverse of f(x) = 3(x + 2)² − 6.

2

Find the inverse of g(x) = −2(x − 1)² + 8 and state its domain.

3

Find the inverse of h(x) = x² − 10x + 21 by completing the square first.

4

f(x) = (x + 1)² − 9 with x ≤ −1. Find f⁻¹(x) and f⁻¹(16).

5

The area of a circle is A(r) = πr². Find r as a function of A, and the radius of a circle with area 50 cm².

Challenge

1

Given f(x) = x², sketch the inverse relation and compare its domain and range with g(x) = √x and h(x) = −√x.

2

A ball's height is h(t) = −5(t − 2)² + 20 metres for 0 ≤ t ≤ 2 (the way up). Find the inverse t(h) and use it to find when the ball is 15 m high on the way up.

3

Explain why f(x) = x² + 4x with domain x ≥ −3 does not have an inverse function, but with domain x ≥ −2 it does.

Check your understanding

Quiz

1.

What is the inverse relation of f(x) = (x + 1)² − 4?

2.

Why is the inverse of y = x² not a function?

3.

The function f(x) = (x − 4)² + 2 is restricted so that its inverse is a function. Which restriction works?

4.

A quadratic has range {y ∈ ℝ | y ≤ 3}. What is the domain of its inverse relation?

5.

f(x) = x², x ≤ 0. What is f⁻¹(x)?

6.

Before finding the inverse of f(x) = x² − 8x + 10, what should you do?

For parents and teachers

Big idea: Expectations A1.5 to A1.7 ask students to find the inverse of a quadratic function numerically, graphically and algebraically, to see that it is not a function unless the domain is restricted, and to compare its domain and range with those of g(x) = √x and h(x) = −√x.

How to help: Ask your teen to explain, in words, why x² = 25 has two answers but a square with area 25 cm² has only one side length. That is the same reasoning as restricting the domain.

Common difficulties:

  • Forgetting the ± after taking a square root.
  • Trying to swap and solve while the quadratic is still in standard form; complete the square first.
  • Sign errors when a < 0 (x − 4 = −(y + 1)² must become 4 − x = (y + 1)²).
  • Choosing the wrong sign after a restriction. Ask: "Which values should the inverse output?"
  • Restricting the domain so that it still includes points on both sides of the vertex. Cutting at the vertex is the usual choice.