How do you find the vertex of the parabola whose equation is #y=x^2-4x+6#?

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Answer 1 · 2016-03-02T08:22:50

the vertex is at (2, 2)

For a parabola of the form

#y=ax^2+bx+c#

The vertex is at #(h,k)#

where #h = (-b)/(2a)#

and #k = f(h)#

In the question posed

a = 1
b = -4
c = 6

Therefore
#h = (-(-4))/(2*1) = 2#

Substituting the value of #h# for #x# in the original formula

#k = 2^2 - 4*2 + 6 = 2#

the vertex is at (2, 2)