SOLUTION: Vertex of x^2+3x+2

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Vertex of x^2+3x+2      Log On


   

Question 495128: Vertex of x^2+3x+2
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Vertex of x^2+3x+2
---
vertex occurs at x = -b/(2a) = -3/2
---
The y-value is (-3/2)^2 + 3(-3/2) + 2 = 9/4-9/2+2 = -1/4
-------------------------
Vertex = (-3/2, -1/4)
==========================
Cheers,
Stan H.
==========

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Vertex of x^2+3x+2
standard form of an equation for a parabola: y=(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
..
y=x^2+3x+2
completing the square
y=(x^2+3x+9/4)+2-9/4
y=(x+3/2)^2-1/4
ans:
coordinates of vertex: (-3/2,-1/4)