What is the vertex form of #y= x^2-3x-28 #?

1 Answer
Feb 1, 2016

#color(blue)"Shortcut method - by sight")#

Given# -> y=x^2-3x-28# .......................................(1)

#y=(x-3/2)^2-3/4-28#

#y=(x-3/2)^2-121/4#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(purple)("Fuller explanation")#

#color(blue)("Step 1")#

Write as#" " y=(x^2-3x)-28#

#color(brown)("Divide the brackets contents by "x". These means that the right")##color(brown)("hand side is no longer equal to "y)#

#y!=(x-3)-28#

#color(brown)("square the brackets")#

#y!=(x-3)^2-28#

#color(brown)("Halve the -3 from "(x-3))#

#y!=(x-3/2)^2-28#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

#color(brown)("Changing the equation so that it does equal "y)#

Let a constant of correction be k then

#y=(x-3/2)^2-28 + k#...................................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

#color(brown)("To find the value of k")#

#color(green)("As equation (1) and equation (2) both equal y we can equate them")# #color(green)("to each other through y")#

Equation (1) = y = Equation (2)

# x^2-3x-28" "=" "(x-3/2)^2-28+k#

# cancel(x^2)-cancel(3x)-cancel(28)" "=" "cancel(x^2)-cancel(3x)+9/4-cancel(28)+k#

#k=-9/4#......................................................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4 - last move!")#

#color(brown)("Bringing it all together to give the final equation")#

Substitute equation (3) into equation (2)

#y=(x-3/2)^2-28 -9/4#.

But #-28-9/4 = -121/4# giving

#color(green)(y=(x-3/2)^2-121/4#.