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 thatThe equation of a vertical parabola in vertex form is equal to[tex]f(x)=a(x-h)^{2}+k[/tex]wherea is a coefficient(h,k) is the vertex of the parabolaIf a > 0 the parabola open upward and the vertex is a minimumIf a < 0 the parabola open downward and the vertex is a maximumIn this problemThe coefficient a must be positive, because we need to find a minimumthereforeCheck the option C and the option DOption Cwe 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)thereforeThe graph of the function [tex]f(x)=\frac{1}{2}x^{2}-4x+5[/tex] has a minimum located at (4,-3)