What is the vertex of the graph of #y=x^2+3x-4#?

1 Answer
Oct 15, 2014

This specific parabola should be transformed into the form

#y = (x - h)^2 + k#

To do that, we need to perform Completing the Square on #x#


Remember that #(ax + b)^2 = a^2x^2 + 2abx + b^2#

For this problem, we have #a = 1 # and #2ab = 3#

#=> 2(1)b = 3#
#=> b = 3/2#


Hence, for #x# to be a perfect square, we need to add #b^2#.
However, we want both sides of the equation to remain equal, so we substract the same value again

#y = (x^2 + 3x + (3/2)^2) - 4 - (3/2)^2#

Simplifying, we have
#y = (x + 3/2)^2 - 25/4#

The parabola's vertex is at #(-3/2, 25/4)#.