Quadratic Functions • Topic 4 of 5

Maximum and Minimum Values

A parabola has a single extreme value at its vertex: a minimum if it opens up (a > 0) and a maximum if it opens down (a < 0). In vertex form y = a(x − h)² + k, that extreme value is k, occurring at x = h. From standard form, find x = −b/(2a) and substitute to get the value. These optimisation questions appear on the SAT as "greatest height," "maximum profit" or "least cost" problems, where you find the vertex and read off the k-value (or the x at which it occurs), so being clear on which the question wants is important.

✅ Solved examples

1. What is the minimum value of y = (x − 2)² + 5?
Opens up; minimum value is k = 5.
2. What is the maximum value of y = −(x + 1)² + 8?
Opens down; maximum value is k = 8.
3. Find the minimum value of y = x² − 4x + 7.
x = 2; y = 4 − 8 + 7 = 3, so the minimum is 3.
4. At what x does y = −x² + 6x attain its maximum?
x = −b/(2a) = 6/2 = 3.

✏️ Practice — try these, take hints as needed

1. What is the minimum value of y = (x − 3)² + 1?
Opens up; min is k.
k = 1.
—
1.
2. What is the maximum value of y = −(x − 2)² + 10?
Opens down; max is k.
k = 10.
—
10.
3. Find the minimum value of y = x² + 2x + 6.
x = −1.
y = 1 − 2 + 6.
—
5.
4. At what x does y = −x² + 8x attain its maximum?
x = −b/(2a).
8/2.
—
x = 4.
5. Does y = 2x² − 4x + 1 have a max or a min?
Check the sign of a.
a = 2 > 0.
—
A minimum.

📝 Topic test — 8 questions

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