Question

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

To find the range of the function \( f(x) = 3x^2 + 6x - 8 \), we first identify its vertex, since this is a quadratic function that opens upwards (the coefficient of \( x^2 \) is positive).

1. **Finding the vertex**: For a quadratic function of the form \( ax^2 + bx + c \), the x-coordinate of the vertex can be calculated using the formula:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 3 \) and \( b = 6 \). Plugging in these values:
\[
x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1
\]

2. **Calculating the y-coordinate of the vertex**: We substitute \( x = -1 \) back into the function to find the corresponding y-coordinate:
\[
f(-1) = 3(-1)^2 + 6(-1) - 8
\]
\[
= 3(1) - 6 - 8
\]
\[
= 3 - 6 - 8 = -11
\]

Thus, the vertex of the parabola is at the point \( (-1, -11) \). Since the parabola opens upwards, the minimum value of \( f(x) \) is \( -11 \), which occurs at the vertex. As \( x \) moves away from the vertex in either direction, \( f(x) \) will increase without bound.

3. **Range of the function**: Therefore, the range of the function \( f(x) \) is all values of \( y \) such that \( y \) is greater than or equal to the minimum value:
\[
\text{Range: } \{ y | y \geq -11 \}
\]

From the options provided, the correct answer is:
\[
\{y | y \geq -11\}
\]