Answer:
The graph of the function [tex]f(x)=\frac{1}{2}x^{2}-4x+5[/tex] has a minimum located at (4,-3)
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]f(x)=a(x-h)^{2}+k[/tex]
where
a is a coefficient
(h,k) is the vertex of the parabola
If a > 0 the parabola open upward and the vertex is a minimum
If a < 0 the parabola open downward and the vertex is a maximum
In this problem
The coefficient a must be positive, because we need to find a minimum
therefore
Check the option C and the option D
Option C
we have
[tex]f(x)=\frac{1}{2}x^{2}-4x+5[/tex]
Convert to vertex form
[tex]f(x)-5=\frac{1}{2}x^{2}-4x[/tex]
Factor the leading coefficient
[tex]f(x)-5=\frac{1}{2}(x^{2}-8x)[/tex]
[tex]f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)[/tex]
[tex]f(x)+3=\frac{1}{2}(x^{2}-8x+16)[/tex]
[tex]f(x)+3=\frac{1}{2}(x-4)^{2}[/tex]
[tex]f(x)=\frac{1}{2}(x-4)^{2}-3[/tex]
The vertex is the point (4,-3) ( is a minimum)
therefore
The graph of the function [tex]f(x)=\frac{1}{2}x^{2}-4x+5[/tex] has a minimum located at (4,-3)
