Let's analyze the given quadratic function to determine its solution set.
The function given is:
[tex]\[ f(x) = 2(x-1)^2 + 4 \][/tex]
This function is in vertex form:
[tex]\[ f(x) = a(x-h)^2 + k \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( h = 1 \)[/tex], and [tex]\( k = 4 \)[/tex].
1. Vertex of the Parabola:
- The vertex form of a quadratic function describes a parabola.
- The vertex of the parabola is at [tex]\( (h, k) = (1, 4) \)[/tex].
2. Direction of the Parabola:
- Since the coefficient [tex]\( a = 2 \)[/tex] is positive, the parabola opens upwards.
3. Minimum Value:
- The minimum value of the function occurs at the vertex because the parabola opens upwards. Thus, the minimum value is [tex]\( f(1) = 4 \)[/tex].
4. Intersection with the x-axis:
- To find if the function has any real solutions (x-intercepts), we set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[
2(x-1)^2 + 4 = 0
\][/tex]
- First, isolate the quadratic term:
[tex]\[
2(x-1)^2 = -4
\][/tex]
- Next, divide both sides by 2:
[tex]\[
(x-1)^2 = -2
\][/tex]
- Since the right side of the equation is negative and the left side is a square of a real number, it means there are no real solutions. Squares of real numbers cannot be negative.
Therefore, the quadratic function [tex]\( f(x) = 2(x-1)^2 + 4 \)[/tex] does not intersect the x-axis and hence has no real solutions.
Now looking at the given options:
A. One real solution
B. Two complex solutions
C. Two real solutions
D. One real solution and one complex solution
Since there are no real solutions, the correct answer must involve complex solutions. Specifically, the solutions are complex because the quadratic equation we derived has imaginary roots when the discriminant is negative as in this case ([tex]\( -2 \)[/tex]). Therefore, the function has two complex solutions.
Thus, the correct answer is:
B. two complex solutions
