What is the vertex form of #y=9x^2-48x+64 #?

1 Answer
Tony B
Feb 27, 2017

You can see a more in-depth build approach example at https://socratic.org/s/aCybisPL

#y=9(x-8/3)^2#

Explanation:

#color(blue)("Preamble")#

If you can do so it is worth committing to memory the standardised form.

Using #y=ax^2+bx+c# as the bases we have the vertex form format of:

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

The extra #k# is a correction that 'gets rid' if the error introduced by squaring the #+b/(2a)# part of #(x+b/(2a))^2#
The #(b/(2a))^2# part is not in the original equation.

Do not forget about the whole bracket being multiplied by a
So to get rid of it we set: #" "a(b/(2a))^2+k=0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Building the vertex form")#

Write as #y=9(x-48/(2(9)))^2+k+64#

#9(-48/18)^2+k=0#

#k=-64#

Thus we have

#y=9(x-8/3)^2-64+64#

#y=9(x-8/3)^2#

Tony B

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