What is the the vertex of #y = (x - 3)^2 + 7x-12 #?

1 Answer
Tony B
Feb 29, 2016

#color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)#

Explanation:

You have to combine these variables before you can do any thing else.

Given:#" "y=(x-3)^2+7x-12#

#y=x^2-6x+9+7x-12#

#y=x^2+x-3#...........................(1)

From this point you can 'complete the square' or do a part version of the process.

I am opting for the part process

Write as:

Consider the coefficient of #+x# in equation ( 1 ), which is 1
Apply #(-1/2)xx1 = -1/2#

#x_("vertex")=-1/2#....... Fast isn't it!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("A word of warning:")# Given the standard form of
#y=ax^2+bx+c# you need to convert this to

#y=a(x^2+b/ax)+c#

In your case #a=1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute #x=-1/2# into equation (1)

#y_("vertex")=(-1/2)^2+(-1/2)-3#

#y_("vertex")=1/4-1/2-3 = -3 1/4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)#

Tony B

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1380 views around the world
You can reuse this answer
Creative Commons License