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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Charles Anctil · Accepted Answer

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

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]}

Christopher Allers · Answer

undefined 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 $.