Question

What is the range of this quadratic function?
y=x2+16x+64

Answer 1

To find the range of the quadratic function \( y = x^2 + 16x + 64 \), we will first rewrite it in vertex form by completing the square.

1. Start with the original equation:
\[
y = x^2 + 16x + 64
\]

2. To complete the square, focus on the \( x^2 + 16x \) part. Take half of the coefficient of \( x \) (which is 16), square it, and add and subtract that square inside the equation:
\[
y = (x^2 + 16x + 64) = (x^2 + 16x + 64) = (x + 8)^2
\]

Since \( 64 = 64 \), we can express:
\[
y = (x + 8)^2
\]

3. The expression \( (x + 8)^2 \) represents a parabola that opens upwards (since the coefficient of \( x^2 \) is positive). The minimum value of this expression occurs when \( (x + 8)^2 = 0 \), which happens when \( x + 8 = 0 \) or \( x = -8 \).

4. At this point, substitute \( x = -8 \) into the function to find the minimum value of \( y \):
\[
y = (-8 + 8)^2 = 0
\]

5. Since the parabola opens upwards, the range of the function starts from the minimum value:
\[
\text{Range of } y = [0, \infty)
\]

Thus, the range of the quadratic function \( y = x^2 + 16x + 64 \) is \( [0, \infty) \).