To find the range of the quadratic function \( y = x^2 + 8x + 16 \), we can start by rewriting it in vertex form.
First, we can complete the square. The given function can be rearranged as follows:
- Take the quadratic expression \( x^2 + 8x \).
- To complete the square, we take half of the coefficient of \( x \) (which is 8), square it, and add and subtract it inside the function. Half of 8 is 4, and \( 4^2 = 16 \).
Rewriting the expression gives:
\[ y = x^2 + 8x + 16 = (x + 4)^2 \]
Now, we have:
\[ y = (x + 4)^2 \]
The vertex of this parabola is at \( (-4, 0) \). Since the coefficient of the squared term is positive (1), the parabola opens upwards.
The minimum value of \( y \) occurs at the vertex, which is \( y = 0 \). As \( x \) moves away from \(-4\) (either towards positive or negative infinity), \( (x + 4)^2 \) increases without bound.
Therefore, the range of the function is:
\[ [0, \infty) \]
In conclusion, the range of the quadratic function \( y = x^2 + 8x + 16 \) is:
\[ \boxed{[0, \infty)} \]