Respuesta :

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  [tex]x^{2}[/tex] , 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 =  [tex](-\frac{b}{2a}, f(-\frac{b}{2a})[/tex]

Where,

  • [tex]a[/tex] is the coefficient of [tex]x^{2}[/tex]
  • [tex]b[/tex] is the coefficient of [tex]x[/tex]

From our equation given, [tex]a=3[/tex]  and  [tex]b=6[/tex]

Now, [tex]x[/tex] coordinate of vertex is  [tex]x=-\frac{b}{2a}\\x=-\frac{6}{(2)(3)}\\x=-\frac{6}{6}\\x=-1[/tex]

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

[tex]3(-1)^{2}+6(-1)-8\\=-11[/tex]


Hence, the range would be all numbers greater than or equal to [tex]-11[/tex]  

Third answer choice is the right one.


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