Let f(x)=x^2-6x+13. what is the vertex form of f(x)? what is the minimum value of f(x)?

The vertex form of a quadratic equation is:

the minimum value of f(x) is

Step-by-step explanation:

Given a quadratic equation of the form then the x coordinate of the vertex is

So for  

Therefore

The x coordinate of the vertex is:

The y coordinate of the vertex is:

By definition the minimum value of the quadratic function is the same as the coordinate of y of its vertex

So the minimum value  is

The vertex form of a quadratic equation is:

Where

a is the main coefficient.

h is the x coordinate of the vertex.

k is the y coordinate of the vertex.

So the vertex form of a quadratic equation is:

a.  

b. The minimum value is 4

Step-by-step explanation:

The given function is:

We add and subtract half the square of the coefficient of x.

This becomes:

The first three terms form a perfect square trinomial.

The function is now in the form:  , where V(h,k) is the vertex.

Therefore the vertex is (3,4).

The minimum value  is the y-value of the vertex, which is 4.

sorry, i can't understand your question. will later if you describe more clearly!

: )

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: