How do you find x-intercepts, coordinates of vertex for parabola #y = x^2 - 4x - 12#?

1 Answer
Alan P.
Apr 11, 2015

#y = x^2-4x-12#

Part 1: x-intercepts
The x-intercepts occur at the points on the function where #y=0#
So, we need to solve
#x^2-4x-12 = 0#
The left side factors fairly easily into:
#(x-6)(x+2)=0#
So solution occur when
#x-6 = 0 rarr x=6#
and
#x+2= 0 rarr x=(-2)#
So the x-intercepts are at #(0,6)# and #(0,-2)#

Part 2: vertex of the parabola
The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to #0#.
The derivative of the given quadratic is
#(dy)/(dx) = 2x -4#
By observation, this is equal to #0# when #x=2#
When #x=2# the original equation becomes
#y = (2)^2 -4(2)-12#
#y = -16#
Therefore the vertex of this parabola is at #(2,-16)#

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