The k is the max/min value: In vertex form a(x − h)² + k, the value k is the minimum (if a > 0) or maximum (if a < 0), reached at x = h.
The squared part is never negative, so it only adds to k.
IB-style question — state the minimum
State the minimum value of f(x) = 2(x − 1)² + 5 and where it occurs.
Step by step
- a = 2 > 0, so the vertex is a minimum.
- Smallest when (x − 1)² = 0, i.e. x = 1.
Final answer
Minimum value 5, at x = 1.
The curve from the worked example, f(x) = 2(x − 1)² + 5: it falls to a lowest point, then rises. The squared part is never negative, so the curve sits lowest where it equals 0 — at x = 1 — giving the minimum value 5. It never dips below 5, so the range is y ≥ 5.
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Value vs point again: The minimum value is k (a number); the minimum point is (h, k).
Re-read which the question wants.
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The vertex height k bounds the range: Every output sits on one side of the vertex height k.
Opens up (a > 0)? k is the lowest point, so y ≥ k.
Opens down (a < 0)? k is the highest, so y ≤ k.
IB-style question — a quadratic's range
State the range of f(x) = (x − 2)² + 3.
Step by step
- Vertex form a(x − h)² + k: vertex (2, 3), and a = 1 > 0 so it opens up.
- The squared part only adds to 3, so the smallest output is 3.
Final answer
Range: y ≥ 3.
f(x) = (x − 2)² + 3 sits lowest at its vertex (2, 3) and fills only the shaded band above — range y ≥ 3.
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Vertex gives the boundary: For a quadratic in vertex form a(x − h)² + k the range is y ≥ k (opens up) or y ≤ k (opens down).
The boundary is always k, never h.
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Standard form? Find the vertex first: If the quadratic is in standard form ax² + bx + c you can't read the range straight off.
Find the vertex first (x = −b/(2a), then substitute), check which way it opens, and the vertex's y-value is the range boundary.
IB-style question — find the range
The function f is defined for x ∈ ℝ by f(x) = 2x² − 4x + 1.
Find the range of f.
Step by step
- a = 2 > 0, so the parabola opens UP — the vertex is the minimum.
- x-coordinate of the vertex.
- Substitute back to get the minimum output.
- Opens up from a minimum of −1, so every output is at or above −1.
Final answer
Range is f(x) ≥ −1 (that is, y ≥ −1).
The curve only ever reaches the shaded band — every output is ≥ −1.
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Two quick routes to the vertex: Use x = −b/(2a) then substitute, or complete the square to a(x − h)² + k and read the vertex (h, k) straight off.
Same answer — pick whichever is faster.