Question

NEED HELP ASAP: Let ​ f(x)=x2−6x+13 ​ . What is the vertex form of f(x)? What is the minimum value of f(x)? Enter your answers in the boxes. Vertex form: f(x)= Minimum value of f(x):

Answer 1

Answer:

I forgot to change the vertex but it's correct

Step-by-step explanation:

Answer 2

The vertex form of  $f(x)=ax^2+bx+c$
$f(x)=a(x-h)^2+k$

where: $h=\dfrac{-b}{2a};\ k=f(h)$

We have:  $f(x)=x^2-6x+13\to a=1;\ b=-6;\ c=13$

substitute:
$h=\dfrac{-(-6)}{2\cdot1}=\dfrac{6}{2}=3\\\\k=f(3)=3^2-6\cdot3+13=9-18+13=4$
The vertex form of f(x):
$f(x)=(x-3)^2+4$
The value of minimum is equal k.
Therefore:  $y_{min}=4$