How do you find the intercept and vertex of #y= -x^2 + 9#?

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Answer 1 · 2016-07-14T06:48:17

Vertex is #(0, 9)#

Its y-intercept is #=9#

Given -

#y=-x^2+9#

It has no #x# term. So we shall have it as -

#y=-x^2+0x+9#

Then it is in the form #y=ax^2+bx+c#

#x# cor-ordinate of the vertex is -

#x=(-b)/(2a)=0/(2xx(-1))=0 #

At #x=0; y=(0^2)+9=9#

Vertex is #(0, 9)#

Its y-intercept is #=9#

Look at the graph