Question

ITS TIMED PLEASE HELP​

Answer 1

Answer:

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ 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

$f(x)=a(x-h)^{2}+k$

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

$f(x)=\frac{1}{2}x^{2}-4x+5$

Convert to vertex form

$f(x)-5=\frac{1}{2}x^{2}-4x$

Factor the leading coefficient

$f(x)-5=\frac{1}{2}(x^{2}-8x)$

$f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x-4)^{2}$

$f(x)=\frac{1}{2}(x-4)^{2}-3$

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)