x = −b/(2a) — the line of symmetry: A parabola is symmetric about a vertical line through its vertex: the axis of symmetry x = −b/(2a). It's also exactly midway between the two x-intercepts.
IB-style question — find the axis
Find the axis of symmetry of y = x² − 6x + 5.
Step by step
- Use x = −b/(2a) with a = 1, b = −6.
Final answer
Axis of symmetry x = 3.
Reach a(x − h)² + k: Completing the square rewrites ax² + bx + c as a(x − h)² + k — which hands you the vertex (h, k) directly. This is the classic "write in the form a(x − h)² + k" exam step.
IB-style question — complete the square
Write x² − 6x + 11 in the form (x − h)² + k.
Step by step
- Halve the x-coefficient (−6 → −3) and square it (9).
- Adjust the constant: 11 = 9 + 2.
Final answer
(x − 3)² + 2, so the vertex is (3, 2).
Half, square, fix the constant: Halve b, square it for the bracket, then add/subtract to keep the constant correct. The vertex is (h, k) — note the sign: (x − 3)² gives h = +3.
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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.
Value vs point again: The minimum value is k (a number); the minimum point is (h, k). Re-read which the question wants.
Start from vertex form, use a point for a: Given the vertex (h, k) and one other point, write y = a(x − h)² + k, substitute the point to find a, and you have the whole quadratic.
IB-style question — find a
A parabola has vertex (2, −3) and passes through (0, 5). Find a in y = a(x − 2)² − 3.
Step by step
- Substitute the point (0, 5).
- Solve for a.
Final answer
a = 2, so y = 2(x − 2)² − 3.