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For real values of \(x\), what is the maximum value of the function \(f(x)=-(x+2)^2+5\)?

Answer and explanation

Correct answer: 5

Since \((x+2)^2\ge 0\), we have \(-(x+2)^2\le 0\), so \(f(x)=-(x+2)^2+5\le 5\). Equality occurs when \(x=-2\), hence the maximum value is 5. Note that \(-2\) is the x-coordinate of the vertex, not the maximum value of the function. Exam tip: for a downward-opening parabola, the y-coordinate of the vertex gives the maximum value.

Related tags

Standard-FunctionsQuadratic-GraphMaximum-ValueGraphs Of Standard FunctionsRelations And FunctionsMathematicsClass 11 Mcq

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