What is the axis of symmetry and vertex for the graph #y = -3x^2 + 12x + 4#?

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Answer 1 · 2018-06-03T00:17:52

aos = 2

vertex = (2,16)

#y = -3x^2 + 12x + 4#

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

In the form #y=ax^2+bx+c# you have:

#a=-3#

#b=12#

#c=4#

Axis of symmetry (aos) is: #aos=(-b)/(2a) = (-12)/(2*-3)= 2#

Remember #y=f(x)#

Vertex is: #(aos, f(aos)) = (2, f(2))#:

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

#f(2) = -3(2)^2 + 12*2 + 4 = 16#

vertex #=(2, 16)#

graph{-3x^2 + 12x + 4 [-16.71, 23.29, -1.6, 18.4]}

Answer 2 · 2018-06-03T00:20:31

Vertex -

#(2,16)#

Axis of symmetry

#x=2#

Given -

#y=-3x^2+12x+4#

Vertex -

#x=(-b)/(2a)=(-12)/(2xx-3)=(-12)/(-6)=2#

At #x=2; y=-3(2^2)+12(2)+4#

#y=-12+24+4=16#

#(2,16)#

Axis of symmetry

#x=2#

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