Question

Vertex of x^2 + 6x + 17

Answer 1

We have:
`f(x)=x^2+6x+17`

Method 1.

Use:
`(*)\ \ \ \ \ (a+b)^2=a^2+2ab+b^2`

and the vertex form
`f(x)=a(x-h)^2+k`

(h, k) - the coordinates of a vertex

`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`

Answer: (-3, 8)

Method 2.

For
`f(x)=ax^2+bx+c` the coordinates of a vertex (h, k) are equal

`h=(-b)/(2a);\ k=f(h)=(-(b^2-4ac))/(4a)`

We have:
`f(x)=x^2+6x+17\to a=1;\ b=6;\ c=17`

Substitute:

`h=(-6)/(2\cdot1)=(-6)/(2)=-3\\\\k=f(-3)=(-3)^2+6(-3)+17=9-18+17=8`

Answer: (-3, 8)