Given
#color(white)("XXX")y=3x^2+12x-6#
This can be converted into vertex form as follows:
#rArr y=3(x^2+4x-2)#
#rArr y=3[x^2+4x-2]#
#rArr y = 3[(x^2+4xcolor(blue)(+4))-2color(blue)(-4)]#
#rArr y=3[(x+2)^2-6]#
#rArr y= 3(x+2)^2-18#
#rArr y= 3(x-color(red)(""(-2)))^2+color(blue)(""(-18))#
which is the vertex form with vertex at #(color(red)(-2),color(blue)(-18))#
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The y-intercept is the value of #y# when #x=color(magenta)0#.
Substituting #color(magenta)0# for #x# in the original equation
#color(white)("XXX")y=3 * color(magenta)0^2+12 * color(magenta) 0 -6=-6#
So the y-intercept is at #y=-6# [or if you prefer at #(0,-6)#]
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The x-intercepts are the values of #x# for points when #y=0#
Using the form derived while determining the vertex form: #y=3(x+2)^2-18#
with #y=color(green)0#
#color(white)("XXX")color(green)0=3(x+2)^2-18#
#color(white)("XXX")rArr color(green)0=(x+2)^2-6#
#color(white)("XXX")rArr (x+2)^2=6#
#color(white)("XXX")rArr x+2=+-sqrt(6)#
#color(white)("XXX")rArr x=-2+-sqrt(6)#
So the x-intercepts are at #x=-2-sqrt(6)# and #x=-2+sqrt(6)# [or if you prefer at #(-2-sqrt(6),0)# and #(-2+sqrt(6),0)#]
