How do you find the vertex of the parabola #y=1/2(x+1)(x-5)#?

1 Answer
Alan P.
Feb 13, 2016

vertex: #(2,-4 1/2)#

Explanation:

Our objective will be to transform the given equation into vertex form:
#color(white)("XXX")y=m(x-color(red)(a))^2+color(blue)(b)# with vertex at #(color(red)(a),color(blue)(b))#
If #y=1/2(x+1)(x-5)#
then
#color(white)("XXX")2y=x^2-4x-5#

#color(white)("XXX")2y= (x^2-4xcolor(green)(+4)) -5 color(green)(-4)# (completing the square)

#color(white)("XXX")2y=(x-2)^2-9#

#color(white)("XXX")y=1/2(x-color(red)(2))^2+color(blue)("("-9/2")")#

which is the vertex form with vertex at #(color(red)(2),color(blue)("("-9/2")"))#

Here's the graph to show that this result is reasonable.
graph{1/2*(x^2-4x-5) [-5.086, 7.4, -5.96, 0.28]}

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