Verified Answer :

nobillionaireNobley

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}

TaeKwonDoIsDead

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.


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