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 ≥ –11
y
TIME REMAINING
22:14
On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 16), has a y-intercept at (0, negative 12), and goes through (2, 0).

What are the x-intercepts of the graph of the function f(x) = x2 + 4x – 12?

(–6, 0), (2,0)
(–2, –16), (0, –12)
(–6, 0), (–2, –16), (2, 0)
(0, –12), (–6, 0), (2, 0)

Answer 1

The range of the function f(x) = 3x² + 6x - 8 is y ≥ -11.

Step-by-step explanation:

To find the range of the function f(x) = 3x² + 6x - 8, we need to determine the set of y-values that the function can output.

Since the parabola opens up and the vertex is the lowest point on the graph, the range of the function is all y-values greater than or equal to the y-coordinate of the vertex. In this case, the vertex is (-1, -11), so the range is y ≥ -11.