Vertex of x^2 + 6x + 17

Question

asked Dec 6, 2019 170k views

1 Answer

Dobes · Answer 1 · 2019-12-11T06:24:29+0000

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)