kase12
kase12 kase12
  • 22-05-2018
  • Mathematics
contestada

What is the vertex form of f(x)?
What is the minimum value of f(x)?

f(x)=x^2-6x+16

Respuesta :

gmany
gmany gmany
  • 23-05-2018
The vertex form of  [tex]f(x)=ax^2+bx+c[/tex]
[tex]f(x)=a(x-h)^2+k[/tex]

where: [tex]h=\dfrac{-b}{2a};\ k=f(h)[/tex]

We have:  [tex]f(x)=x^2-6x+16\to a=1;\ b=-6;\ c=16[/tex]

substitute:
[tex]h=\dfrac{-(-6)}{2\cdot1}=\dfrac{6}{2}=3\\\\k=f(3)=3^2-6\cdot3+16=9-18+16=7[/tex]
The vertex form of f(x):
[tex]f(x)=(x-3)^2+7[/tex]
The value of minimum is equal k.
Therefore:  [tex]y_{min}=7[/tex]

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