A quadratic function attains its minimum value of 5 at x=1 and holds a value of 6 at x=0. Find the value of the function at x=5

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Solution

The correct option is B
21

Standard form of a quadratic equation = f(x) = $a x^{2}$ +bx+c=0

Substitute x=0
Given that f(x)=c=6
Substitute x=1
a+b+c=5
a+b =-1................(1)
The minimum value of a function occurs at x= $- (\frac{b}{2 a})$ , which is given as 1
Thus, 2a=-b
Substituting back in eqn (1) a=1 and b=-2
$f (x) = x^{2} - 2 x + 6$

Thus, at x=5, f(x)=21. Option (b)

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