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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $x$ is a real, then minimum value of $x^2-8x+17$ is:
Solution
Given
$f(x)=x^2-8x+17$
Complete the square:
$x^2-8x+17=(x^2-8x+16)+1$
$=(x-4)^2+1$
Since $(x-4)^2 \ge 0$
Minimum value occurs when $(x-4)^2=0$
So minimum value = $1$
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