What is the vertex form of #y=7x^2-9x-32#?

1 Answer
Jan 17, 2016

#y_("vertex form")=7(x-9/14)^2-977/28#

Explanation:

Given: #y=7x^2-9x-32#......................(1)

Write as:

#y=7(x^2-9/7x)-32#

Now write as

#y=7(x-[1/2xx9/7])^2-32 color(blue)(+"correction")#

#y=7(x-9/14)^2-32color(blue)(+"correction")#..........................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider #7(x-9/14)^2#
This gives: #7(x^2-9/7x+81/196)#
We need the #7(x^2-9/7x)# but the #7(+81/196)# is an extra value we need to get rid of. This is why we have a correction. In this case the correction value is:#color(blue)(7( -81/196)=-81/28)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So equation (2) becomes:

#y=7(x-9/14)^2-32color(blue)(+(-81/28))#..........................(2_a)

#y=7(x-9/14)^2-977/28#

#("So " x_("vertex")=(-1)xx(-9/14)color(white)(..)=+9/14)#