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

Question

Ethan Adkins · Accepted Answer

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

Savannah Anderson · Answer

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#

Isaiah Anthony · Answer

undefined The vertex form of the quadratic equation $y = -3x^2 - 5x + 9$ is $y = -3(x + \frac{5}{6})^2 + \frac{161}{12}$.