What is the vertex form of #y=-3x^2 - 5x + 9 #?

2 Answers
Apr 11, 2018

#y=-3(x+5/6)^2+133/12#

Explanation:

#y=-3[x^2+5/3]+9#
#y=-3[(x+5/6)^2-25/36]+9#
#y=-3(x+5/6)^2+25/12+9#
#y=-3(x+5/6)^2+133/12#

Apr 11, 2018

Given: #y=-3x^2-5x+9#

Write as: #y=-3(x^2+5/3x)+9" ".................Equation(1)#

Consider the #(color(green)(x^2+5/3x)) # part

We need to make this a #ul("'perfect square'")# but in 'forcing' it to do this we introduce a value that is not in the original equation. To correct this we have to turn it into 0 by subtraction or addition as appropriate by the same amount. Rather like #a+2# being changed to #(a+2) +3-3#

#color(green)(-3[x^2+5/3x] color(white)("ddd")->color(white)("ddd")-3[(x+5/(2xx3))^2]#

#color(green)(color(white)("dddddddddddddd")->color(white)("ddd")-3[x^2+5/3xcolor(red)(color(white)(.)ubrace(+(5/6)^2))])#
#color(white)("ddddddddddddddddddddddddddddddddd.d")color(red)(uarr)#
#color(white)("dddddddddddddddddddddddd")color(red)("The introduced error")#

Substitute this into #Equation(1)#

#color(green)(y=-3(x^2+5/3x)+9#

#color(white)("dddddddddddddddd")color(red)("The error")#
#color(white)("ddddddddddddddddd.d")color(red)(darr)#
#color(green)(y=ubrace(-3[x^2+5/3xcolor(red)(color(white)(.)+obrace((5/6)^2))])+color(blue)(k)+9)" " k# is the correction
#color(white)("ddddddddddd.d")color(green)(darr)#
#color(green)(y=color(white)("ddd")-3(x+5/6)^2color(white)("ddddd")+color(blue)(k)+9#

The whole error is #color(red)((-3)xx(5/6)^2)#

#color(green)(y=color(white)("ddd")-3(x+5/6)^2+color(blue)([3xx(5/6)^2]) +9)#

#color(white)()#

#y=-3(x+5/6)^2 +133/12#