Maximum and minimum at the vertex
Write the function in vertex form: y = a(x − p)² + q. The vertex is (p, q). The square (x − p)² is never negative.
- If a > 0: the smallest value is q at x = p. This is the minimum. There is no maximum (the curve goes up for ever).
- If a < 0: the largest value is q at x = p. This is the maximum. There is no minimum.
From standard form y = ax² + bx + c, the vertex is at x = −b ÷ 2a. Put this x back in the function to get the value.
Maximum and minimum on a limited domain
Often x is allowed only in an interval p ≤ x ≤ q (the domain). Then the answer can change, because the vertex may be outside it. Use this method:
- Find the vertex x-value, x = −b ÷ 2a.
- Check: is it between the two ends?
- Find f at both ends. Find f at the vertex only if it is inside.
- The largest of these values is the maximum. The smallest is the minimum.
If the vertex is outside, the curve only rises or only falls inside the domain, so the extreme values are at the ends. If the vertex is inside, it gives the minimum (a > 0) or the maximum (a < 0), and the other extreme is at an end.
Applied problems: area, profit and height
Steps to solve a word problem:
- Choose the unknown as x and note what values are allowed (for example 0 < x < 10 for a length).
- Write the quantity to be made largest or smallest as a quadratic in x.
- Find the vertex, and check it is in the allowed range.
- Answer in words with the unit.
Fence problem. A rope of length 20 m makes a rectangle with sides x and 10 − x. Area A = x(10 − x) = −x² + 10x. Vertex at x = 5, so A = 25 m². Among rectangles of the same perimeter, the square has the largest area.
Height problem. A ball is thrown so that h = 20t − 5t² (h in m, t in s). Vertex at t = 2 s, h = 20 m. This is the highest point.
Try it
Take a 20 cm piece of string. Make a rectangle with sides 1 and 9, then 3 and 7, then 5 and 5, and measure each area on grid paper. Write the areas in a row. You will see them rise and then fall, with the square on top. Now slide the 3D band and see which ends give the max and min.
Key formulas and definitions
- y = a(x − p)² + q: vertex (p, q)
- Vertex of y = ax² + bx + c: x = −b / 2a
- a > 0: minimum q at x = p; a < 0: maximum q at x = p
- On p ≤ x ≤ q: compare f(p), f(q) and f(vertex) if the vertex is inside
- Rectangle with perimeter 2s: area x(s − x), maximum s²/4 at x = s/2
Worked examples
1. Find the minimum of y = x² − 6x + 5.
Vertex x = −(−6)/(2·1) = 3. y = 9 − 18 + 5 = −4. Since a = 1 > 0, the minimum is −4 at x = 3.
2. Find the maximum of y = −2x² + 8x + 1.
Vertex x = −8/(2·(−2)) = 2. y = −8 + 16 + 1 = 9. Since a < 0, the maximum is 9 at x = 2.
3. Find the maximum and minimum of y = x² − 4x + 1 for 0 ≤ x ≤ 1.
Vertex x = 2 is outside [0, 1]. So compare ends: f(0) = 1, f(1) = 1 − 4 + 1 = −2. Maximum 1 at x = 0, minimum −2 at x = 1.
4. Find the maximum and minimum of y = x² − 4x + 1 for 1 ≤ x ≤ 5.
Vertex x = 2 is inside. f(2) = −3. Ends: f(1) = −2, f(5) = 25 − 20 + 1 = 6. Maximum 6 at x = 5, minimum −3 at x = 2.
5. Find the maximum and minimum of y = −x² + 2x for 2 ≤ x ≤ 4.
Vertex x = 1 is outside [2, 4]. f(2) = −4 + 4 = 0, f(4) = −16 + 8 = −8. Maximum 0 at x = 2, minimum −8 at x = 4.
6. A rectangle has perimeter 40 m. Find the largest possible area.
The two sides are x and 20 − x. A = x(20 − x) = −x² + 20x. The vertex is x = 10, so A = 100 m². The best rectangle is a 10 m by 10 m square.
7. A ball is thrown up, with height h = 20t − 5t² metres after t seconds. Find the highest point.
Vertex t = −20/(2·(−5)) = 2 s. h = 40 − 20 = 20 m. The ball reaches 20 m after 2 s.
Common mistakes
- Forgetting to check the domain. On 0 ≤ x ≤ 1 the vertex x = 2 is outside, so it must not be used.
- Giving the x-value when the question asks for the maximum value (which is y), or the other way round.
- Saying a parabola with a > 0 has a maximum. On all real x it has only a minimum.
- Writing a wrong sign in x = −b/2a, for example using b instead of −b.