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]

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