How do you find the vertex of # y = 10x – 3x^2#?

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Answer 1 · 2017-05-02T15:48:07

Using a 'sort of' cheat. Not really! It is part of the process of completing the square.

Vertex#->(x,y)=(5/3,25/3)#

Note that as #-3x^2# is negative the graph is of shape #nn#

Write as #y=-3x^2+10x+0#

This is now in the standardised form of

#y=ax^2+bx+c#

Change this to:

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

Note that #axxb/a=b#

#x_("vertex")=(-1/2)xxb/a#

#x_("vertex")=(-1/2)xx10/(-3) = +10/6 = 5/3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute #x_("vertex")=+5/3# giving:

#y_("vertex")=-3(5/3)^2+10(5/3)#

#y_("vertex")=-75/9+50/3#

#y_("vertex")=-75/9+150/9#

#y_("vertex")=+75/9=8 1/3 ->25/3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Vertex#->(x,y)=(5/3,25/3)#

Tony B