Given:
#y=-3x^2-4x+6# is a quadratic equation in standard form,
#y=ax^2-bx+c#,
where:
#a=-3#, #b=-4#, and #c=6#
Axis of Symmetry: value of the vertical line #x# in which the parabola will be divided into two equal halves; also the #x# value for the vertex.
#x=(-b)/(2a)#
#x=(-(-4))/(2*-3)#
#x=-4/6#
#x=-2/3# #larr# #x# axis of symmetry and #x# value of the vertex.
Vertex: minimum or maximum point, #(x,y)#, of a parabola. If #a>0#, the vertex is the minimum point and the parabola will open upward. If #a<0#, the vertex is the maximum point and the parabola will open downward.
To determine the #y# value of the vertex, substitute #-2/3# for #x# and solve for #y#. Recall that a whole number, #n# is a fraction of #n/1#.
#y=-3/1(-2/3)^2-4(-2/3)+6#
#y=-3/1(4/9)+8/3+6#
The least common denominator is #9#. Multiply #8/3# by #color(teal)(3)/color(teal)(3)#, and #6/1xxcolor(magenta)(9)/color(magenta)9# to make all numbers have the same (common) denominator.
#y=-12/9+8/3xxcolor(teal)(3)/color(teal)(3)+6/1xxcolor(magenta)(9)/color(magenta)9#
#y=-12/9+24/9+54/9#
#y=((-12+24+54))/9#
#y=66/9#
#y=22/3#
Vertex: #(-2/3,22/3)# #larr# maximum point of the parabola
Approximate vertex: #(-0.667,7.33)#
Range: all real numbers for #y# that are part of the parabola.
Range: #{yinRR:color(white)(.)y<=22/3}#
graph{y=-3x^2-4x+6 [-10, 10, -5, 5]}
