Example: suppose we had #a^2-b^2# this is the same as #(a-b)(a+b)#
Note that #(x-2)(x+2)# is of the same structure
Multiply out the brackets
#" "3(x^2-2^2) = 3x^2+12" "larr 3(-2)^2=3xx4=12#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Observation 1")#
The #3x^2# is positive so the general shape of the graph is #uu# thus there is a #color(green)("minimum")#.
If it had been #-3x^2# then the general shape would have been #nn#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Observation 2")#
This is a quadratic equation so the #color(green)("axis of symmetry is the "x" coordinate of the vertex")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the y intercept")#
Consider the standard form of #y=ax^2+bx+c#
In the question's equation ther is no #bx# term but there is a #c# term in that #c=+12#
#color(green)(y_("intercept")=c=+12)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Observation 3") ->("Determine x vertex")#
The #bx# term in #y=ax^2+bx+c# moves the graph left or right from the y axis. As there is no # bx # term the graph is symmetrical about the y-axis.
#color(green)("Thus "x_("vertex")-> x=0)" "color(red)(larr" axis of symmetry")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Observation 4") ->("Determine y vertex")#
From above the graph is symmetrical about the y-axis. Consequently y-vertex must be at the point where the graph crosses the y axis, which is:
#color(green)(y_("intercept")=c=+12)" "color(red)(larr" " y_("vertex")#
To prove my point; substitute #x=0# into #y=3x^2+12#
#color(green)(y_("vertex")=3(0^2)+12=12)#
#color(green)("Vertex"->(x,y)->(0,12))#
