To find the range of the quadratic function [tex]\( f(x) = 2x^2 - 16x + 28 \)[/tex], we need to follow these steps:
1. Determine the direction of the parabola:
Since the coefficient of [tex]\( x^2 \)[/tex] is positive (2 in this case), the parabola opens upwards. This implies that the function has a minimum value at the vertex.
2. Find the vertex of the parabola:
The vertex form of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] occurs at [tex]\( x = -\frac{b}{2a} \)[/tex].
Given:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -16 \)[/tex]
- [tex]\( c = 28 \)[/tex]
Calculate the x-coordinate of the vertex:
[tex]\[
x = -\frac{b}{2a} = -\frac{-16}{2 \cdot 2} = \frac{16}{4} = 4
\][/tex]
3. Find the y-coordinate of the vertex:
Substitute [tex]\( x = 4 \)[/tex] into the function to find [tex]\( f(4) \)[/tex]:
[tex]\[
f(4) = 2(4)^2 - 16(4) + 28 = 2 \cdot 16 - 64 + 28 = 32 - 64 + 28 = -4
\][/tex]
Therefore, the vertex of the parabola is at [tex]\( (4, -4) \)[/tex].
4. Determine the range of the function:
Since the parabola opens upwards, the y-coordinate of the vertex represents the minimum value of [tex]\( f(x) \)[/tex]. Thus, the minimum value of the function is [tex]\( -4 \)[/tex].
The range of the function is all y-values greater than or equal to this minimum value. In interval notation, the range is:
[tex]\[
\boxed{[-4, \infty)}
\][/tex]
