If f(x)=x^2+ax+b has a minima at x=3 whose va | vedclass

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(A) Given the function $f(x) = x^2 + ax + b$. To find the critical points,we calculate the first derivative: $f'(x) = 2x + a$. Setting $f'(x) = 0$,we

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-$6$,$14$

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(A) Given the function $f(x) = x^2 + ax + b$. To find the critical points,we calculate the first derivative: $f'(x) = 2x + a$. Setting $f'(x) = 0$,we get $x = -a/2$. Since the second derivative $f''(x) = 2 > 0$,the function has a local minimum at $x = -a/2$. According to the problem,the minimum occurs at $x = 3$. Therefore,$-a/2 = 3$,which implies $a = -6$. Given that the minimum value of the function is $5$ at $x = 3$,we substitute these values into the original function: $f(3) = (3)^2 + a(3) + b = 5$. Substituting $a = -6$: $9 + (-6)(3) + b = 5$. $9 - 18 + b = 5$. $-9 + b = 5$. $b = 14$. Thus,the values are $a = -6$ and $b = 14$.

If $f(x)=x^2+ax+b$ has a minima at $x=3$ whose value is $5$,then the values of $a$ and $b$ are respectively.

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