Question

Determine the Range of the given function k(x) = 6x ^ 2 + 12x + 5
All real numbers
1 <= x
- 4 <= y
- 1 <= y

Answer 1

To determine the range of the function \( k(x) = 6x^2 + 12x + 5 \), we can start by rewriting the function in vertex form, which helps us find the minimum or maximum value.

First, we can complete the square:

1. Start with the quadratic function:
\[
k(x) = 6x^2 + 12x + 5
\]

2. Factor out the coefficient of \( x^2 \) from the first two terms:
\[
k(x) = 6(x^2 + 2x) + 5
\]

3. To complete the square inside the parentheses, we take half of the coefficient of \( x \) (which is \( 2 \)), square it (getting \( 1 \)), and then add and subtract that inside the parentheses:
\[
k(x) = 6(x^2 + 2x + 1 - 1) + 5
\]
\[
= 6((x + 1)^2 - 1) + 5
\]
\[
= 6(x + 1)^2 - 6 + 5
\]
\[
= 6(x + 1)^2 - 1
\]

Now we have the function in vertex form:
\[
k(x) = 6(x + 1)^2 - 1
\]

From this form, we can see that the vertex of the parabola described by this function is at the point \((-1, -1)\). Since the coefficient of the squared term (\(6\)) is positive, the parabola opens upwards.

The minimum value of \( k(x) \) occurs at the vertex, which is \( -1 \). Therefore, the range of \( k(x) \) is all values greater than or equal to the minimum value.

Thus, the range of the function \( k(x) \) is:
\[
\boxed{[-1, \infty)}
\]