How do you find the vertex and intercepts for #y = x^2 - 4x + 12#?

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Answer 1 · 2017-07-18T11:10:38

Vertex is at #(2,8)#, y- intercept is #y=12 # or at #(0,12)# ,
x-intercept is absent.

#y=x^2-4x+12 or y= x^2-4x+4+8 # or

#y= (x-2)^2 + 8#, Comparing with standard equation of vertex

form, # y= a(x-h)^2+k ; (h,k)# being vertex , we find here

#h=2 , k =8#. So vertex is at #(2,8)#. y-intercept can be found by

putting #x=0# in the equation #y=x^2-4x+12 :. y = 12#

y- intercept is #y=12 # or at #(0,12)# . x-intercept can be found by

putting #y=0# in the equation #y=x^2-4x+12# or

#x^2-4x+12=0 ;(ax^2+bx+c=0) a=1 , b=-4 ,c=12 #

Discriminant #D= b^2-4ac= 16-48=-32 :. D<0# .

Since #D<0# the roots are complex in nature , so there is no

x-intercept. x-intercept is absent.

graph{x^2-4x+12 [-40, 40, -20, 20]} [Ans]