How do I use the vertex formula to determine the vertex of the graph for #-3x^2+12x-9#?

2 Answers
May 18, 2015

Vertex form: #f(x) = a(x+b/(2a))^2 + f(-b/(2a))#

x of vertex:# x =( -b/(2a) )= -4/-2 = 2#

y of vertex: f(-2) = -12 + 24 - 9 = 3

Vertex form: #f(x) = -3(x - 2)^2 + 3#

Check by developing:

# f(x) = -3(x^2 - 4x + 4) + 3 = -3x^2 + 12x - 9#. OK

May 18, 2015

For #y=ax^2+bx+c#, the vertex formula tells us that the #x#-coordinate of the vertex is #-b/(2a)#

For #y=-3x^2+12x-9#, we have: #a=-3# and #b=12#

Therefore, the vertex formula gives us #x# coordinate is

#-12/(2(-3)) = -12/(-6) = 1(-2) = 2#

We still need the #y# coordinate of the vertex, so use the equation to find it:

#y = -3(2)^2+12(2)-9 = -12+24-9 = 3#

The vertex is #(2,3)#

graph{y=-3x^2+12x-9 [-10, 10, -5, 5]}