SOLUTION: sketch the graph of f(x)=2x^2-8x+11, it is a parabola with its lowest (vertex) in which Quadrant? Use window of -10 to 10 both x and y

Algebra ->  Equations -> SOLUTION: sketch the graph of f(x)=2x^2-8x+11, it is a parabola with its lowest (vertex) in which Quadrant? Use window of -10 to 10 both x and y      Log On


   

Question 907758: sketch the graph of f(x)=2x^2-8x+11, it is a parabola with its lowest (vertex) in which Quadrant? Use window of -10 to 10 both x and y
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
the vertex form of a Parabola opening up(a>0) or down(a<0),
y=a%28x-h%29%5E2+%2Bk where(h,k) is the vertex and x = h is the Line of Symmetry
f(x)=2x^2-8x+11 = 2(xhighlight_green%28-2%29)^2 highlight_green%28-8%29 + 11 = 2%28x-2%29%5E2+%2B3
Completing Square for ax^2+bx+c, Note: b/2a = %28-8%29%2F%282%2A2%29= -2 and highlight_green%28-8%29= (-)2(-2)^2
f(x)=2(x-2)^2 + 3 ..... V(2,3) Quadrant I