Question

What is the range of this quadratic function?
y=8x2+32x+24

Answer 1

To find the range of the quadratic function \( y = 8x^2 + 32x + 24 \), we first need to identify its vertex, since the vertex will provide information about the minimum or maximum point of the parabola.

1. **Identify the coefficients**: The given quadratic function is in the standard form \( y = ax^2 + bx + c \), where:
- \( a = 8 \)
- \( b = 32 \)
- \( c = 24 \)

2. **Locate the vertex**: The x-coordinate of the vertex of a quadratic function can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Plugging in the values for \( b \) and \( a \):
\[
x = -\frac{32}{2 \cdot 8} = -\frac{32}{16} = -2
\]

3. **Find the corresponding y-coordinate**: Now, we substitute \( x = -2 \) back into the quadratic function to find the minimum value of \( y \):
\[
y = 8(-2)^2 + 32(-2) + 24
\]
\[
= 8 \cdot 4 - 64 + 24
\]
\[
= 32 - 64 + 24
\]
\[
= -8
\]

4. **Determine the direction of the parabola**: Since the coefficient \( a \) (which is 8) is positive, the parabola opens upwards. This means that the vertex at \( (-2, -8) \) is the minimum point of the function.

5. **Find the range**: The range of the quadratic function is all values of \( y \) greater than or equal to the minimum value at the vertex. Thus, the range is given by:
\[
y \geq -8
\]
In interval notation, the range is:
\[
[-8, \infty)
\]

Therefore, the range of the quadratic function \( y = 8x^2 + 32x + 24 \) is \(\boxed{[-8, \infty)}\).