Question

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

Answer 1

Answer:

Range is {y | y ≥ –11}

Step-by-step 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 =  $(-\frac{b}{2a}, f(-\frac{b}{2a})$

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  $x=-\frac{b}{2a}\\x=-\frac{6}{(2)(3)}\\x=-\frac{6}{6}\\x=-1$

$y$ 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.

Answer 2

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|y ≥ –11}