Question

What is the range of this quadratic function?
y=x^2+8x+16

Answer 1

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:

1. Take the quadratic expression \( x^2 + 8x \).
2. 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:

\[
\boxed{[0, \infty)}
\]