Question

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

Answer 1

To determine the range of the quadratic function \( y = -x^2 - 8x - 12 \), we can first rewrite the function in vertex form. This will help us find the vertex, which gives us the maximum (or minimum) value of the function.

1. The given function is:
\[
y = -x^2 - 8x - 12
\]

2. To convert it into vertex form, we can complete the square. Start by factoring out the leading coefficient from the \( x \) terms:
\[
y = -1(x^2 + 8x) - 12
\]

3. Now, complete the square for the expression \( x^2 + 8x \):
- Take half of the coefficient of \( x \) (which is \( 8 \)), square it, and add and subtract that value:
- Half of \( 8 \) is \( 4 \), and \( 4^2 = 16 \).
\[
y = -1(x^2 + 8x + 16 - 16) - 12
\]
\[
y = -1((x + 4)^2 - 16) - 12
\]
\[
y = -1(x + 4)^2 + 16 - 12
\]
\[
y = -1(x + 4)^2 + 4
\]

4. Now the function is in vertex form \( y = -1(x + 4)^2 + 4 \). The vertex of this parabola is at \( (-4, 4) \).

5. Since the coefficient of \( (x + 4)^2 \) is negative, the parabola opens downward, meaning it has a maximum point at the vertex.

6. The maximum value of \( y \) is \( 4 \). As the parabola opens downward, the function can take any value less than or equal to \( 4 \).

Thus, the range of the function is:
\[
\{y | y \leq 4\}
\]