How do you find the intercepts, vertex and graph #f(x)=x^2-9x+9#?

1 Answer
Narad T.
Oct 30, 2016

The intercepts are #(0,9)#, #((9+sqrt45)/2,0)# and #((9-sqrt45)/2,0)#
And the vertex is #(9/2,-45/4)#

Explanation:

Let's rewrite #f(x)# in the vertex form
So, #f(x)=x^2-9x+9=x^2-9x+81/4+9-81/4#
#f(x)=(x-9/2)^2-45/4#

So the vertex is #(9/2,-45/4)#
The y intercepts are #x=0##=>##y=9#

The x intercepts are when #y=0#
#f(x)=x^2-9x+9=0#
so we calculate #Delta=b^2-4ac=81-4*9*1=45#
and we get #x=(9+-sqrt45)/2#
So the x intercepsts are #((9+sqrt45)/2,0)# and #((9-sqrt45)/2,0)#
You can also use #f(x)=0#
#(x-9/2)^2-45/4=0#
#(x-9/2)^2=45/4#
#(x-9/2)=+-sqrt45/2#
#x=9/2+-sqrt45/2=(9+-sqrt45)/2#

graph{x^2-9x+9 [-28.86, 28.85, -14.43, 14.43]}

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