Question

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

Answer 1

The vertex form of a quadratic equation is:

`f(x) = (x-3)^2+4`

the minimum value of f(x) is
`y=4`

Explanation:

Given a quadratic equation of the form
`f (x) = ax ^ 2 + bx + c` then the x coordinate of the vertex is

`x=-(b)/(2a)`

So for
`f(x)=x^2-6x+13`

`a=1\\b=-6\\c=13\\`

Therefore

The x coordinate of the vertex is:

`x=-((-6))/(2(1))`

`x=3`

The y coordinate of the vertex is:

`f(3)=(3)^2-6(3)+13`

`y=f(3)=4`

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
`y=4`

The vertex form of a quadratic equation is:

`f(x) = a(x-h)^2+k`

Where

a is the main coefficient.
`a=1`

h is the x coordinate of the vertex.
`h=3`

k is the y coordinate of the vertex.
`k=4`

So the vertex form of a quadratic equation is:

`f(x) = (x-3)^2+4`

Answer 2

a.
`f(x)=(x-3)^2+4`

b. The minimum value is 4

Explanation:

The given function is:
`f(x)=x^2-6x+13`

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

`f(x)=x^2-6x+3^2-3^2+13`

This becomes:
`f(x)=x^2-6x+9-9+13`

The first three terms form a perfect square trinomial.

`f(x)=(x-3)^2+4`

The function is now in the form:
`f(x)=a(x-h)^2+k`, 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.