Question

find the range of x^2+8x+19

Answer 1

To find the range of the quadratic function \( f(x) = x^2 + 8x + 19 \), we can rewrite it in vertex form by completing the square.

Starting with the original equation:

\[
f(x) = x^2 + 8x + 19
\]

1. **Complete the square**:
- Take the coefficient of \( x \) (which is 8), divide it by 2, and square it:
\[
\left(\frac{8}{2}\right)^2 = 16
\]

2. **Rewrite the function**:
\[
f(x) = (x^2 + 8x + 16) + 19 - 16
\]
\[
f(x) = (x + 4)^2 + 3
\]

Now, the function is in the form \( f(x) = (x+4)^2 + 3 \).

3. **Determine the vertex**:
- The vertex of the quadratic \( (x + 4)^2 + 3 \) is \( (-4, 3) \).
- Since the quadratic coefficient (the coefficient of \( x^2 \)) is positive, the parabola opens upwards.

4. **Find the range**:
- The minimum value of \( f(x) \) occurs at the vertex, which is \( 3 \).
- Since the parabola opens upwards, \( f(x) \) can take any value greater than or equal to \( 3 \).

Thus, the range of the function \( f(x) = x^2 + 8x + 19 \) is:

\[
\boxed{[3, \infty)}
\]