Restrict a quadratic inverse and reflect its graph
Choose one branch so reversal remains a function, then exchange coordinates across the line y = x.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Written practice is self-checked; show your reasoning and domain restrictions.
Before you begin
- Complete a square and find a quadratic's vertex and range.
- Rearrange an equation to make a variable the subject and understand principal square roots.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why does f(x) = (x − 1)² + 2 on all real x fail to have an inverse function?
What you will learn
- Use a domain restriction to make a quadratic one-one and select its inverse branch.
- Sketch a function and its inverse as reflections in y = x with exchanged domain and range.
The domain selects the square-root sign
The rule f(x) = (x − 1)² + 2 on all real x is not one-one. Inputs 0 and 2 both produce 3, so reversing all its pairs would send input 3 to two outputs. Restrict f to x ≥ 1 to use the right branch. Its range is y ≥ 2, and every output is then attained by just one permitted input.
Solve y − 2 = (x − 1)². The domain x ≥ 1 makes x − 1 non-negative, so choose x = 1 + √(y − 2). Thus f⁻¹(x) = 1 + √(x − 2), with domain x ≥ 2 and range y ≥ 1. If instead f were restricted to x ≤ 1, the inverse would use the minus branch 1 − √(x − 2), with range y ≤ 1. A ± expression would give two outputs and would not be an inverse function.
Reflect the permitted graph, not the full parabola
A point (a, b) on a one-one function becomes (b, a) on its inverse. Exchanging coordinates reflects the graph in the line y = x; label that mirror line on a sketch. Use equal unit lengths on the two axes if drawing a geometric reflection. The original domain and range exchange roles, including their open or closed endpoints.
For the right-branch example, points (1, 2), (2, 3) and (3, 6) reflect to (2, 1), (3, 2) and (6, 3). The original vertex becomes the inverse's endpoint. The displayed graph window may cut off part of either curve; it does not create a new upper domain limit. Do not draw the forbidden left branch and then claim its complete reflection is this inverse.
Pause and explain
For f(x) = (x − 1)² + 2 on x ≥ 1, which inverse is valid?
Worked example
Find the inverse of f(x) = (x − 1)² + 2 on x ≥ 1, stating its domain and range.
Show the worked solution
- The original range is y ≥ 2; the right branch is one-one.
- Solve y − 2 = (x − 1)². Since x − 1 ≥ 0, use x = 1 + √(y − 2).
- Rename the inverse input and exchange the original output and input sets.
Answer f⁻¹(x) = 1 + √(x − 2), domain x ≥ 2, range y ≥ 1.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the inverse of f(x) = (x + 2)² + 1 for x ≥ −2, including its input and output restrictions.
Give me a hint
The original bracket x + 2 is non-negative on the chosen branch.
Compare my reasoning
- The original range is y ≥ 1.
- From y − 1 = (x + 2)², use x + 2 = √(y − 1), giving x = −2 + √(y − 1).
- Rename the input; the inverse domain is x ≥ 1 and its range is y ≥ −2.
f⁻¹(x) = −2 + √(x − 1), domain x ≥ 1, range y ≥ −2.
Look for these in your work
- Select the branch from the original domain.
- Exchange both restrictions correctly.
Find the inverse of f(x) = 3(x − 2)² − 4 for x ≤ 2, with its domain and range.
Give me a hint
Divide by 3 before taking a square root, then select the left branch.
Compare my reasoning
- The original range is y ≥ −4; rearrange (y + 4)/3 = (x − 2)².
- Because x − 2 ≤ 0, solve x = 2 − √((y + 4)/3).
- The inverse domain is x ≥ −4 and its range is y ≤ 2.
f⁻¹(x) = 2 − √((x + 4)/3), domain x ≥ −4, range y ≤ 2.
Look for these in your work
- Keep the division by 3 inside the square root.
- Select the minus branch and exchange domain with range.
For f(x) = (x − 1)² + 2 on x ≥ 1, use the points (1, 2), (2, 3) and (3, 6) to plan a sketch of f⁻¹. State the mirror line and the inverse's endpoint.
Give me a hint
Swap each point's two coordinates and retain the permitted branch.
Compare my reasoning
- Reflect the points to (2, 1), (3, 2) and (6, 3).
- Draw and label the mirror line y = x, using equal axis unit lengths for the reflection.
- The original vertex (1, 2) becomes inverse endpoint (2, 1); the inverse is defined for x ≥ 2 and continues beyond this sample window.
Inverse points (2, 1), (3, 2), (6, 3); mirror line y = x; endpoint (2, 1).
Look for these in your work
- Exchange both coordinates of every point.
- Mark the mirror line and explain the endpoint restriction.
Common mistakes
- The chosen domain determines whether the inverse uses a plus or minus square-root branch.
- An inverse function cannot use ± to give two outputs for a single input.
- Reflect only the allowed branch and label y = x as the mirror line.
What can you explain now?
What is the inverse of the same rule f(x) = (x − 1)² + 2 when its domain is x ≤ 1?
Compare with the explanation
f⁻¹(x) = 1 − √(x − 2), domain x ≥ 2, range y ≤ 1.
On the left branch x − 1 ≤ 0, so the inverse must select the negative square root before adding 1.
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Make it stick
Find the two possible single-branch inverses of a completed-square quadratic. Explain why only one is selected for each stated original domain.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.