Complete the square to rewrite y = x2 - 6x + 16 in vertex form. then state
whether the vertex is a maximum or minimum and give its coordinates.

minimum (3,7)

Step-by-step explanation:

y = x^2 - 6x + 16

Take the coefficient of the x term  -6

Divide it by 2   -6/2 =-3

Then square it  ( -3)^2 =9

Add that to the equation (remember if we add it we must subtract it)

y = x^2 - 6x +9  -9+ 16

y = (x^2 - 6x +9)  -9+ 16

The term inside the parentheses is x+b/2 which is x+ -3  or x-3 quantity squared

y = (x-3)^2 +7

This is in vertex form

y = a(x-h)^2 +k  where (h,k) is the vertex

(3,7) is the vertex

Since a=1, it is positive, so it opens upward and the vertex is a minimum

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.