How do you find the vertex and the intercepts for # f(x) = x^2 - 6x + 3#?

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Answer 1 · 2017-06-11T10:59:14

Vertex is at #(3,-6)#, y-intercept is at #(0,3)#
x-intercepts are at #(3+sqrt6,0) and (3-sqrt6,0) #

#f(x)=x^2-6x+3 = (x-3)^2 -9+3= (x-3)^2 -6#. Comparing with standard equation #y=a(x-h)^2+k ; (h,k)# being vertex,we find here #h=3,k= -6#.

So vertex is at #(3,-6)#

y-intercept can be found by plug in #x=0# in the equation.

#f(x) = 0-0+3 =3 #. So y-intercept is at #(0,3)#

x-intercept can be found by plug in #f(x) =0# in the equation.

#(x-3)^2-6=0 or (x-3)^2=6 or (x-3) = +-sqrt6 or x= 3+-sqrt6#

So x-intercepts are at #(3+sqrt6,0) and (3-sqrt6,0) #

graph{x^2-6x+3 [-40, 40, -20, 20]} [Ans]