Respuesta :

gmany

The minimum value of a square function that is open up is in the vertex.

The formula of a vertex:

[tex]f(x)=ax^2+bx+c\\\\V(p;\ q)\\\\p=\dfrac{-b}{2a};\ q=f(p)=\dfrac{-(b^2-4ac)}{4a}[/tex]

We have:

[tex]g(x)=x^2-6x-12\\\\a=1,\ b=-6;\ c=-12[/tex]

a = 1 > 0 therefore the parabola is open up.

[tex]p=\dfrac{-(-6)}{2\cdot1}=\dfrac{6}{2}=3\\\\q=g(3)=3^2-6(3)-12=9-18-12=-21[/tex]

Answer: The minimum value of function g(x) is equal -21 for x = 3.

raphealnwobi

The minimum value of the function g(x) =x² - 6x - 12 is at -21.

Polynomial is an expression consisting of only the operations of addition, subtraction, multiplication of variables.

Polynomials are classified by degree as linear, quadratic, cubic, etc.

Given the polynomial, g(x) =x²-6x-12, the minimum value is at g'(x) = 0. Hence:

g'(x) = 2x - 6

2x - 6 = 0

2x = 6

x = 3

g(3) = (3)² - 6(3) - 12 = -21

The minimum value of the function g(x) =x² - 6x - 12 is at -21.

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