On a coordinate plane, a parabola opens up. it goes through (negative 3, 0), has a vertex at (negative 1, negative 11), and goes through (1, 0). what is the range of the function f(x) = 3x2 + 6x – 8? {y|y ≥ –1} {y|y ≤ –1} {y|y ≥ –11} {y|y ≤ –11}

3 is greater then both of those numbers

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places

it would be a on edgenuit y

the diagram below is your final answer