We have: [tex]f(x)=x^2+6x+17[/tex]
Method 1.
Use: [tex](*)\ \ \ \ \ (a+b)^2=a^2+2ab+b^2[/tex]
and the vertex form [tex]f(x)=a(x-h)^2+k[/tex]
(h, k) - the coordinates of a vertex
[tex]x^2+6x+17=x^2+2\cdot x\cdot3+17=\underbrace{x^2+2\cdot x\cdot3+3^2}_{(*)}-3^2+17\\\\=(x+3)^2+8\to h=-3;\ k=8[/tex]
Answer: (-3, 8)
Method 2.
For [tex]f(x)=ax^2+bx+c[/tex] the coordinates of a vertex (h, k) are equal
[tex]h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[/tex]
We have: [tex]f(x)=x^2+6x+17\to a=1;\ b=6;\ c=17[/tex]
Substitute:
[tex]h=\dfrac{-6}{2\cdot1}=\dfrac{-6}{2}=-3\\\\k=f(-3)=(-3)^2+6(-3)+17=9-18+17=8[/tex]
Answer: (-3, 8)
