How do you find the vertex and the intercepts for #y=x^2 - 9x#?

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Answer 1 · 2017-06-28T06:23:51

Vertex #(4.5, -20.25)#
Intercepts #(0,0); (0,9)#

Given -

#y=x^2-9x#

Vertex

#x=(-b)/(2a)= (-(-9))/(2xx1)=9/2=4.5#
At #x=4.5#
#y=(4.5)^2-9(4.5)#
#y=20.25-40.5=-20.25#

Vertex #(4.5, -20.25)#

Intercepts
At #y=0#

#x^2-9x=0#
#x(x-9)=0#
#x=0#
#x=9#

Intercepts #(0,0); (0,9)#

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