How do you find the axis of symmetry, vertex and x intercepts for #y=-x^2+6x-5#?

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Answer 1 · 2017-02-15T16:42:35

The axis of symmetry is #x=3#
The vertex is #=(3,4)#
The y-intercept is #=(0,-5)#
The x-intercepts are #(1,0)# and #(5,0)#

Let's complete the squares

#y=-x^2+6x-5#

#y=-(x^2-6x)-5#

#y=-(x^2-6x+9)-5+9#

#y=-(x-3)^2+4#

This is the vertex form of the equation

The axis of symmetry is #x=3#

The vertex is when #x=3#, #=>#, #y=4#

The y-intercept is when #x=0#, #=>#, #y=-5#

The x intercepts are when #y=0#

#-(x-3)^2+4=0#

#(x-3)^2=4#

#x-3=2# and #x-3=-2#

#x=5# and #x=1#

graph{-x^2+6x-5 [-11.25, 11.25, -5.625, 5.625]}