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If f(x)=x^2+2x-8, what is the minimum value of f(x)?

Answer and explanation

Correct answer: -9

The governing concept is the minimum value of an upward-opening quadratic function. Complete the square: f(x) = x² + 2x − 8 = (x + 1)² − 9. For every real value of x, the square (x + 1)² is at least zero, and it reaches zero when x = −1. Therefore the least possible value of f(x) is 0 − 9 = −9. The parabola opens upward because the coefficient of x² is positive, so this vertex value is the global minimum. Option A is correct. Option B misses the constant adjustment, option C does not follow from the completed-square form, and option D incorrectly assumes that the minimum must be zero.

Related tags

Real-Valued-FunctionsMinimum-ValueQuadratic-FunctionReal Valued FunctionsTheir Domain And RangeReal Valued Functions Their Domain And RangeRelations And FunctionsMathematicsClass 11 Mcq

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