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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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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Answer 1 · 2018-01-16T00:07:52

$a = -6$ 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]}

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

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Explanation:

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