What is the vertex of #y=-x^2 +12x - 4 #?

1 Answer
Nov 23, 2015

#x=6# I will let you solve for #y# by substation.

#color(brown)("Look at the explanation. It shows you a short cut!")#

Explanation:

Standard form: #y=ax^2+bx_c=0 color(white)(....)#Where

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=-1#
#b=12#
#c=-4#

#color(blue)(~~~~~~~~~~~~ " Short Cut " ~~~~~~~~~~~~)#

#color(brown)("Change into format of " y= ax^2+bx+c " into:")#
#color(brown)(y=a(x^2 +b/ax+c/a) color(white)(xxx) ->color(white)(.....) (-1)(x^2-12x+4))#

#color(blue)("THE TRICK!")# # color(white)(....)color(green)(x_("vertex")= (-1/2) (b/a) = (-1/2)(-12)=+6)#

#color(blue)(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~)#

#color(red)(" To demonstrate the point - 'The long way round !'")#

The factors of 4 will not produce the sum of 12 so use the formula

The vertex #x# will be the mean of the two #x's# that are a solution the standard form

#a=-1#
#b=12#
#c=-4#

Thus

#x= (-(12)+-sqrt(12^2-(4)(-1)(-4)))/(2(-1))#

#x=+6+-(sqrt(144-16))/(-2)#

#x= + 6 +- (sqrt(128))/(-2)#

#x= 6 +- (sqrt(2xx64))/(-2)#

#x= 6 +- (8sqrt(2))/(-2)#

#x= 6 +- (-4sqrt(2))#

Tony B

The mean point is:

#x_("vertex")= ((6-4sqrt(2)) +(6+4sqrt(2)))/2 =6#

Substitute #x_("vertex")=6# into the original equation to find the value of #y_("vertex")#