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

$h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}$

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

Substitute:

$h=\dfrac{-6}{2\cdot1}=\dfrac{-6}{2}=-3\\\\k=f(-3)=(-3)^2+6(-3)+17=9-18+17=8$

Answer: (-3, 8)