How do you write #f(x) = (x-3)^2 + 17# in standard form?

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Answer 1 · 2016-07-23T10:48:37

#y=x^2-6x+26#

The given form is that of 'Vertex form' or 'Completing the Square'

We need to change this so that it is in the form of #y=ax^2+bx+c#

Write as #y=(x-3)^2+17#

This is the same as:

#y=color(blue)((x-3)) color(brown)((x-3))+17#

Multiply out the brackets

#y=color(brown)(color(blue)(x)(x-3)color(blue)(-3)(x-3))+17 #

#y=x^2-3x" "-3x+9" "+17#

#y=x^2-6x+26#