What is the the vertex of y = (x - 3)^2 + 7x-12 y=(x3)2+7x12?

1 Answer
Feb 29, 2016

color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)Vertex(x,y)(12,314)

Explanation:

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

Given:" "y=(x-3)^2+7x-12 y=(x3)2+7x12

y=x^2-6x+9+7x-12y=x26x+9+7x12

y=x^2+x-3y=x2+x3...........................(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+x in equation ( 1 ), which is 1
Apply (-1/2)xx1 = -1/2(12)×1=12

x_("vertex")=-1/2xvertex=12....... Fast isn't it!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(brown)("A word of warning:")A word of warning: Given the standard form of
y=ax^2+bx+cy=ax2+bx+c you need to convert this to

y=a(x^2+b/ax)+cy=a(x2+bax)+c

In your case a=1a=1
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute x=-1/2x=12 into equation (1)

y_("vertex")=(-1/2)^2+(-1/2)-3yvertex=(12)2+(12)3

y_("vertex")=1/4-1/2-3 = -3 1/4yvertex=14123=314
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)Vertex(x,y)(12,314)

Tony B