Sure, let's work through this problem step by step:
Given the quadratic function:
[tex]\[ f(x) = 2x^2 - 16x + 25 \][/tex]
### (a) Determine whether the quadratic function has a maximum value or a minimum value. Then find this value.
First, observe the coefficient of [tex]\( x^2 \)[/tex], which is 2. Since the leading coefficient (2) is positive, the parabola opens upwards. Therefore, the quadratic function has a minimum value.
To find this minimum value, we need to find the vertex of the parabola. The x-coordinate of the vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex] and [tex]\( b = -16 \)[/tex]. Plugging these values into the vertex formula:
[tex]\[ x = -\frac{-16}{2 \cdot 2} = \frac{16}{4} = 4 \][/tex]
Now, substitute [tex]\( x = 4 \)[/tex] back into the function to find the minimum value:
[tex]\[ f(4) = 2(4)^2 - 16(4) + 25 \][/tex]
[tex]\[ f(4) = 2(16) - 64 + 25 \][/tex]
[tex]\[ f(4) = 32 - 64 + 25 \][/tex]
[tex]\[ f(4) = -7 \][/tex]
So, the minimum value of the function is [tex]\( -7 \)[/tex].
### (b) Find the range of [tex]\( f \)[/tex].
Since the quadratic function opens upwards and has a minimum value of [tex]\( -7 \)[/tex] at [tex]\( x = 4 \)[/tex], the range of [tex]\( f \)[/tex] includes all values greater than or equal to [tex]\( -7 \)[/tex].
Therefore, the range of [tex]\( f \)[/tex] is:
[tex]\[ [-7, \infty) \][/tex]
### Summary:
#### (a) The function [tex]\( f \)[/tex] has a minimum value.
[tex]\[
\text{The minimum value of } f(x) \text{ is } -7.
\][/tex]
#### (b) The range of [tex]\( f \)[/tex] is:
[tex]\[
[-7, \infty)
\][/tex]
