how do i find the minimum value or maximum value of function
x^2-12x+9?

2. Originally Posted by Rose Wanjohi
how do i find the minimum value or maximum value of function
x^2-12x+9?
Rewrite it in vertex form:
$\displaystyle y = a(x - h)^2 + k$.

If the leading coefficient is positive, then the vertex (h, k) is the minimum point; if the leading coefficient is negative, then the vertex (h, k) is the maximum point.

So, we complete the square to rewrite the equation in vertex form:
$\displaystyle x^2-12x+9$
\displaystyle \begin{aligned} &= x^2 - 12x \;{\color{red}+\;36}\;+ 9\; {\color{red}-\;36} \\ &= (x - 6)^2 - 27 \end{aligned}

The leading coefficient was positive (a = 1), so the point (6, -27) is the minimum of the parabola.

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3. Originally Posted by Rose Wanjohi
how do i find the minimum value or maximum value of function
x^2-12x+9?
Hello: The formula for the minimum or the
the maximum point by the function f(x) = a x 2 +bx +c is :
(-b/2a; f(-b/2a))
In this exercice : a =1 , b= -12 c =9
I'have : (-(-12)/2 ,f(6))
f(6) = 36-12(6)+9=-27 Conclusion : A ( 6 , -27 )