What is the range of this quadratic function?

y=x2+12x+20

1 answer

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 \([-16, \infty)\).