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

y
y ≤ –1
y
y ≤ –11

User Nada
by
8.2k points

2 Answers

1 vote
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
User FussinHussin
by
7.4k points
2 votes

Answer:

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.


User Wpakt
by
8.6k points

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