How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y = -3x^2 + 12x - 8#?

1 Answer
Alan P.
Jun 13, 2015

Convert the equation into vertex form. Use the vertex and parabolic form to determine the axis of symmetry. Set one variable to #0# to determine the intercept of the other.

Explanation:

Conversion to Vertex Form
#color(white)("XXXX")# vertex form is #y=m(x-a)^2+b# with vertex at #(a,b)#
#y=-3x^2+12x-8#
#rarr##color(white)("XXXX")# #y = (-3)(x^2-4x) -8#
#rarr##color(white)("XXXX")# #y =(-3)(x^2-4x+4) -8 +12#
#rarr##color(white)("XXXX")# #y = (-3)(x-2)^2+4#
The vertex is at #(2,4)#

Axis of Symmetry
The equation is a parabola in standard position (vertical axis) opening downward. So the axis of symmetry is a vertical line running through the vertex:
#color(white)("XXXX")# #x=2#.

Intercepts
#color(white)("XXXX")# y-intercept
Set #x=0# in equation to determine the y-intercept:
#y = 3(0)^2+12(0)-8#
#y=-8#
#color(white)("XXXX")# the y-intercept is #(-8)#

#color(white)("XXXX")# x-intercept
Set #y=0# and solve for #x# to determine the x-intercepts
#0 = 3x^2+12x-8#
Use the quadratic formula #x = (-b+-sqrt(b^2-4ac))/(2a)#
to determine x-intercepts

#x " intercepts" = (-12+-sqrt(144 +96))/6#

#color(white)("XXXX")# #= -2+- sqrt(240)/6#

#color(white)("XXXX")# #= -2+- (2sqrt(15))/3#

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