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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1 Answer
Jan 16, 2018

Answer:

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

Explanation:

We have:

# f(x) = x^2 + ax + b #

We know that #y=f(x)# passes through #(3,9)#, thus:

# 9 = 3^2+3a+b=> 3a+b = 0 .... [A] #

We also require a minimum at this coordinate, (we know that it will have a minimum as we have positive coefficient of #x^2# so differentiating wrt #x# we have:

# f'(x) = 2x+a #

A critical points occurs when:

# f'(3)=0 => 6+a = 0 => a =-6 #

Substituting into Eq [A] we get:

# -18 + b =0 => b = 18 #

Hence:

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

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