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

1 Answer
May 1, 2017

y intercept: #color(red)(-2)#
vertex at #color(red)(""(x,y)=(2,-6))#

Explanation:

The y-intercept is the value of #y# when #x=0# (that is for point on the Y-axis)
for the given equation
#color(white)("XXX")y=x^2-4x-2#
replacing #x# with #0# gives
#color(white)("XXX")y=0^2-4**0-2 = -2#

The easiest way to find the vertex is to convert the given equation into "vertex form": #y=m(x-a)^2+b# with vertex at #(a,b)#
#color(white)("XXX")y=x^2-4x-2#

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

#color(white)("XXX")y=1(x-2)^2-6#
which is the vertex form with vertex at #(2,-6)#

For verification purposes, here is teh graph of the original equation:
graph{x^2-4x-2 [-4.636, 7.854, -6.12, 0.12]}