For a real number \(x\), what is the minimum value of \(p(x)=x^2-6x+10\)?
Answer and explanation
Correct answer: 1
Completing the square gives \(p(x)=x^2-6x+10=(x-3)^2+1\). Since \((x-3)^2\geq 0\), we have \(p(x)\geq 1\), and the minimum value 1 occurs at \(x=3\). Option 3 is the \(x\)-coordinate of the vertex, not the minimum value. Exam tip: Rewrite a quadratic as \((x-a)^2+b\); when the square term has a positive coefficient, its minimum value is \(b\).
Frequently asked questions
What is the correct answer to this question?
1
Why is this the correct answer?
Completing the square gives \(p(x)=x^2-6x+10=(x-3)^2+1\). Since \((x-3)^2\geq 0\), we have \(p(x)\geq 1\), and the minimum value 1 occurs at \(x=3\). Option 3 is the \(x\)-coordinate of the vertex, not the minimum value. Exam tip: Rewrite a quadratic as \((x-a)^2+b\); when the square term has a positive coefficient, its minimum value is \(b\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Polynomials in one variable.
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