Question

What is the range of this quadratic function?
y=–x2–8x–12
{x|x ≤ -1/2}
{x|x ≥ -25/4}
{x|x ≥ -1/2}
all real numbers

Answer 1

To determine the range of the quadratic function \( y = -x^2 - 8x - 12 \), we can start by rewriting it in vertex form by completing the square.

The standard form of a quadratic function is given as:

\[
y = ax^2 + bx + c
\]

For the function \( y = -x^2 - 8x - 12 \):

1. Identify \( a = -1 \), \( b = -8 \), and \( c = -12 \).
2. Complete the square:

First, factor out -1 from the terms involving \( x \):

\[
y = - (x^2 + 8x) - 12
\]

Next, we complete the square for \( x^2 + 8x \). The coefficient of \( x \) is \( 8 \); take half of that and square it:

\[
\left( \frac{8}{2} \right)^2 = 16
\]

Now, we can write:

\[
x^2 + 8x = (x + 4)^2 - 16
\]

Substituting back, we have:

\[
y = - ((x + 4)^2 - 16) - 12
\]

This simplifies to:

\[
y = - (x + 4)^2 + 16 - 12
\]
\[
y = - (x + 4)^2 + 4
\]

The vertex form of the quadratic is now \( y = - (x + 4)^2 + 4 \). The vertex of this parabola is at the point \( (-4, 4) \).

Since the coefficient of \( (x + 4)^2 \) is negative, this parabola opens downward. Therefore, the maximum value of \( y \) is 4 (at the vertex), and it decreases without bound.

Thus, the range of the function is:

\[
(-\infty, 4]
\]

Given the possible answers listed in your question, none of them seem to directly represent this range. However, based on our calculation, the correct range is \( y \in (-\infty, 4] \).