What is the the vertex of #y = (x+6)(x+4) -x+12#?

2 Answers
Jun 23, 2016

#y_{min} = 63/4 # at #x = - 9/2#

Explanation:

#y = (x+6)(x+4) -x+12#
#y = x^2 + 10x + 24 -x+12#
#y = x^2 + 9x + 36#
#y = (x + 9/2)^2 - 81/4 + 36#
#y = (x + 9/2)^2 + 63/4#

#y_{min} = 63/4 # at #x = - 9/2#

Desmos

Jun 23, 2016

The vertex is #(-9/2;63/4)#

Explanation:

let's rewrite the equation in the equivalent form:

#y=x^2+4x+6x+24-x+12#

#y=x^2+9x+36#

Then let's find the vertex coordinates by the following:

#x_V=-b/(2a)#

where a=1; b=9

so

#x_V=-9/2#

and

#y_V=f(-9/2)#

that's

#y=(-9/2)^2+9(-9/2)+36#

#y=81/4-81/2+36#

#y=(81-162+144)/4#

#y=63/4#