Question

What is the range of this quadratic function?
y=x2–x–12
choices
{y|y ≤ 1/2}
{y|y ≤1/2}
{y|y ≥-49/4}
all real numbers

Answer 1

To find the range of the quadratic function \( y = x^2 - x - 12 \), we first need to determine the vertex of the parabola defined by this quadratic equation.

The standard form of a quadratic function is \( y = ax^2 + bx + c \). In this case:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -12 \)

1. **Find the x-coordinate of the vertex**: The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \).

\[
x = -\frac{-1}{2 \cdot 1} = \frac{1}{2}
\]

2. **Find the y-coordinate of the vertex**: Substitute \( x = \frac{1}{2} \) back into the function to find the y-coordinate.

\[
y = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 12
\]
\[
= \frac{1}{4} - \frac{1}{2} - 12
\]
\[
= \frac{1}{4} - \frac{2}{4} - \frac{48}{4}
\]
\[
= \frac{1 - 2 - 48}{4} = \frac{-49}{4}
\]

The vertex of the parabola is at \( \left( \frac{1}{2}, -\frac{49}{4} \right) \).

Since the coefficient of \( x^2 \) (which is \( a = 1 \)) is positive, the parabola opens upwards. Therefore, the minimum value of \( y \) occurs at the vertex, which is \( -\frac{49}{4} \).

3. **Determine the range**: As the parabola opens upwards and its minimum value is \( -\frac{49}{4} \), the range of the function is:

\[
y \geq -\frac{49}{4}
\]

In set notation, this is expressed as:

\[
\{ y \mid y \geq -\frac{49}{4} \}
\]

Among the given choices, the correct range of the quadratic function is:

\(\{ y | y \geq -\frac{49}{4} \}\)