# The curve of #y=f(x)# where #f(x) = x^2 + ax + b # has a minimum at #(3,9)#. Find #a# and #b#?

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(Question Restore: portions of this question have been edited or deleted!)

(Question Restore: portions of this question have been edited or deleted!)

# a = -6# and#b=18# making# f(x) = x^2 -6x +18 #

We have:

A critical points occurs when:

Substituting into Eq [A] we get:

Hence:

Which we confirm graphically: graph{x^2 -6x +1 [-5, 10, -10, 5]}

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To find ( a ) and ( b ), we need to use the fact that the curve has a minimum at ( (3, 9) ). Since the curve has a minimum, its derivative at that point must be zero.

Given ( f(x) = x^2 + ax + b ), its derivative is ( f'(x) = 2x + a ). At the minimum point ( (3, 9) ), the derivative ( f'(3) ) should be zero.

So, we have:

[ f'(x) = 2x + a ] [ f'(3) = 2(3) + a = 0 ]

Solving for ( a ), we get:

[ 6 + a = 0 ] [ a = -6 ]

Now, to find ( b ), we can use the fact that ( f(3) = 9 ):

[ f(3) = 3^2 + (-6)(3) + b = 9 ]

Solving for ( b ), we get:

[ 9 - 18 + b = 9 ] [ b = 18 ]

So, ( a = -6 ) and ( b = 18 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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