Question

What is the range of this quadratic function?
y=x2+12x+20

Answer 1

To find the range of the quadratic function \( y = x^2 + 12x + 20 \), we can start by rewriting the quadratic in vertex form. This will allow us to easily identify the vertex, which gives us the minimum (or maximum) value of the function.

1. **Completing the square:**

- Start with the original function:
\[
y = x^2 + 12x + 20
\]

- Rearranging the first two terms:
\[
y = (x^2 + 12x) + 20
\]

- To complete the square, take half of the coefficient of \( x \) (which is 12), square it and then add it inside the parentheses:
\[
\left(\frac{12}{2}\right)^2 = 36
\]

- Add and subtract 36 inside the parentheses:
\[
y = (x^2 + 12x + 36 - 36) + 20
\]
\[
y = (x + 6)^2 - 36 + 20
\]
\[
y = (x + 6)^2 - 16
\]

2. **Identifying the vertex:**

The quadratic is now in vertex form \( y = (x + 6)^2 - 16 \), where the vertex is at the point \( (-6, -16) \).

3. **Determining the range:**

Since the coefficient of \( (x + 6)^2 \) is positive, the parabola opens upwards. Thus, the minimum value of \( y \) occurs at the vertex. Therefore, the minimum value of \( y \) is \( -16 \).

Given that the parabola opens upwards, the range of the function starts from this minimum value:

\[
\text{Range: } y \geq -16
\]

In interval notation, the range is:
\[
\text{Range: } [-16, \infty)
\]

Thus, the range of the quadratic function \( y = x^2 + 12x + 20 \) is \([-16, \infty)\).