What is the vertex of the equation, y= -3x^2+2x-7? a. (1/3, -20/3) b. (-1/3, -20/3) c. (-1/3, -19/3) d. (1/3, -19/3)

A

Given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 ), then

The x-coordinate of the vertex is

= -

y = - 3x² + 2x - 7 is in standard form

with a = - 3, b = 2 and c = - 7, hence

= - =

Substitute this value into the equation for y-coordinate

y = - 3 ()² + 2 ( ) - 7

  = - + - = -

vertex = ( , - ) → A

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

honestly bro aint no one got time to answer these dumb question just do it p***y

step-by-step explanation: