Find a Quadratic Maximum Without Confusing Time and Height

For times from zero to six, the height model is negative two t squared plus twelve t plus five. We need the maximum height, not the time when it occurs.

The negative leading coefficient makes the parabola open downward. The vertex time is negative b divided by two a: negative twelve divided by negative four, or three.

Substitute three into the original model: negative eighteen plus thirty six plus five equals twenty three. The maximum height is twenty three meters. Three is the time, not the height.

Completing the square gives negative two times the quantity t minus three squared, plus twenty three. That negative square cannot be positive. The vertex is permitted, and both endpoints have height five, confirming the maximum of twenty three.

The vertex formula gives t = −12/(2 × −2) = 3. Direct substitution gives h(3) = −18 + 36 + 5 = 23. Completing the square gives h(t) = −2(t − 3)² + 23, so the value cannot exceed 23. The allowed interval contains t = 3, and both endpoints have height 5.

For all real x, what is the maximum value of −3(x − 4)² + 29? At what x does it occur?

Answer: The maximum value is 29 at x = 4. The square is nonnegative, so its product with −3 is at most zero. At x = 4 the square vanishes. Keep the input x = 4 separate from the requested output 29.

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