Maximum and Minimum Values

6.6: Maximum and Minimum Values. Original editorial SAT preparation with three worked examples, ten flashcards, and thirty practice questions. Educator-review draft.

Lesson 6 of 83 parts Practice Flashcards

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Minimum in vertex form

Objective: Find the least value of an upward-opening quadratic.

Prerequisite skills: vertex form and completing the square. Every real square is nonnegative. For a > 0, a(x − h)² + k is at least k and reaches k at x = h. That conclusion assumes h is in the allowed domain. Common error: reporting the input h when asked for the minimum output. Restrictions and interpretation: A vertex minimum requires a > 0; a maximum requires a < 0; restrict a contextual domain when stated.

Formulas

a > 0 ⇒ minimum of a(x − h)² + k is k

Worked examples

Try itWhat is the minimum value of f(x) = 3(x − 23)² + 28 for real x?Show answer

The square is at least 0 and the coefficient 3 is positive. The lowest output occurs at x=23, making the square zero; f=28. Therefore the requested answer is 28.

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Minimum in vertex form

Objective: Find the least value of an upward-opening quadratic.

Formulas

a > 0 ⇒ minimum of a(x − h)² + k is k

Worked examples

Try itWhat is the minimum value of f(x) = 3(x − 23)² + 28 for real x?Show answer

The square is at least 0 and the coefficient 3 is positive. The lowest output occurs at x=23, making the square zero; f=28. Therefore the requested answer is 28.

Maximum in vertex form

Objective: Find the greatest value of a downward-opening quadratic.

Multiplying a nonnegative square by a negative number makes a nonpositive quantity. Adding k gives an output no greater than k. Equality occurs when the square is zero. Common error: calling a downward-opening parabola a minimum. Restrictions and interpretation: A vertex minimum requires a > 0; a maximum requires a < 0; restrict a contextual domain when stated.

Formulas

a < 0 ⇒ maximum of a(x − h)² + k is k

Worked examples

Try itWhat is the maximum value of g(x) = −2(x + 23)² + 34?Show answer

Because −2(x+23)² ≤ 0, g(x) ≤ 34. Equality is attained at x=-23; the maximum output is 34. Therefore the requested answer is 34.

Optimization from standard form

Objective: Complete a square or use the vertex formula to optimize.

For ax² + bx + c, the vertex input is −b/(2a). Substitute it into the function or complete the square to obtain the extreme value. Distinguish the optimizing input from the extreme output. Common error: giving the vertex x-coordinate as the maximum value. Restrictions and interpretation: A vertex minimum requires a > 0; a maximum requires a < 0; restrict a contextual domain when stated. Desmos and method choice: Use Desmos for arithmetic or a graph/table check when that reduces work. Enter parentheses explicitly and verify the requested quantity and units. Use the algebraic steps shown here for an exact result; a rounded graph display is not proof of exact equality.

Formulas

h = −b/(2a), extreme output = f(h)

Worked examples

Try itFor real x, what is the maximum of P(x) = −x² + 52x + 43?Show answer

Complete the square: P(x) = −(x − 26)² + 719. The negative square is at most 0, so the maximum is 719, attained at x=26. Therefore the requested answer is 719.

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