Question

On a coordinate plane, a parabola opens up. It goes through (negative 3, 0), has a vertex at (negative 1, negative 11), and goes through (1, 0).
What is the range of the function f(x) = 3x2 + 6x – 8?

{y|y ≥ –1}
{y|y ≤ –1}
{y|y ≥ –11}
{y|y ≤ –11}

Answer 1

The parabola has vertex (−1, −11) and opens upward (coefficient 3 > 0). Completing the square:
f(x) = 3(x+1)^2 − 11,
so the minimum value is −11. Therefore the range is {y | y ≥ −11}.