Question

What is the range of the function f(x) = 3x2 + 6x – 8?

y
y ≤ –1
y
y ≤ –11

Answer 1

3x^2 + 6x - 8 = 3(x^2 + 2x - 8/3) = 3(x^2 + 2x + 1 - 8/3 - 1) = 3(x + 1)^2 + 3(-8/3 - 1) = 3(x + 1) - 11
Vertex = (-1, -11)
Range = y ≥ –11

Answer 2

Range is y ≥ –11

Explanation:

This is quadratic equation.

A quadratic equation's range can be found if we find the vertex.

For quadratic equations that have a positive number in front of
`x^(2)` , it is upward opening and thus all the numbers greater than or equal to the minimum value of vertex is the range.

The formula for vertex of a parabola is:

Vertex =
`(-(b)/(2a), f(-(b)/(2a))`

Where,

• 
  `a` is the coefficient of
  `x^(2)`
• 
  `b` is the coefficient of
  `x`

From our equation given,
`a=3` and
`b=6`

Now,
`x` coordinate of vertex is
`x=-(b)/(2a)\\x=-(6)/((2)(3))\\x=-(6)/(6)\\x=-1`

`y` coordinate of the vertex IS THE MINIMUM VALUE that we want. We get this by plugging in the
`x` value [
`x=-1` ] into the equation. So we have:

`3(-1)^(2)+6(-1)-8\\=-11`

Hence, the range would be all numbers greater than or equal to
`-11`

Third answer choice is the right one.