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

Answer:

The vertex form of a quadratic equation is:

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

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

Step-by-step explanation:

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

$x=-\frac{b}{2a}$

So for  $f(x)=x^2-6x+13$

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

Therefore

The x coordinate of the vertex is:

$x=-\frac{(-6)}{2(1)}$

$x=3$

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

Answer:

a.  $f(x)=(x-3)^2+4$

b. The minimum value is 4

Step-by-step 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.