How do you find the vertex and intercepts for #y = 3x^2 - 6x - 4#?

1 Answer
Feb 21, 2016

vertex: #(1,-7)#
y-intercept: #(-4)#
x-intercepts: #1+-sqrt(7/3)#

Explanation:

One way to find the vertex is to convert the given equation into vertex form:
#color(white)("XXX")y=3x^2-6x-4#

#color(white)("XXX")y=3(x^2-2x) -4#

#color(white)("XXX")y=3(x^2-2xcolor(green)(+1))color(green)(-3)-4#

#color(white)("XXX")y=3(x-1)^2+(-7)#
which is the vertex form for a parabola with vertex at #(1,-7)#

the y-intercept can be found by setting #x=0# in the original equation:
#color(white)("XXX")y=3(0)^2-6(0)-4=-4#

The x-intercepts are a bit more work, but can be found by setting #y=0# in any one of the equations and solving for #x#.
I find this easiest with the vertex form above:
#color(white)("XXX")0=3(x-1)^2-7#

#color(white)("XXX")3(x-1)^2=7#

#color(white)("XXX")(x-1)^2=7/3#

#color(white)("XXX")x-1=+-sqrt(7/3)#

#color(white)("XXX")x=1+-sqrt(7/3)~~2.53" or "-0.53#

A graph of the original equation might help show that these results are reasonable:
graph{3x^2-6x-4 [-4.13, 11.67, -7.17, 0.73]}