What is the vertex of #y=-x^2+6#?

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Answer 1 · 2017-03-06T17:45:38

#(0,6)#

Consider the standardised form of #y=ax^2+bx+c#

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

#x_("vertex")=(-1/2)xxb/a" "->" "(-1/2)xx0/(-1) = 0#

The y-intercept# = c = 6#

As there is no #bx# term in #y=-x^2+6" "# the axis of symmetry is the y-axis. So the vertex is at #(x,y)=(0,6)#

As the #x^2# term is negative then the general shape of the curve is #nn#

Tony B