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

Jul 18, 2017

Vertex is at $\left(2 , 8\right)$, y- intercept is $y = 12$ or at $\left(0 , 12\right)$ ,
x-intercept is absent.

#### Explanation:

$y = {x}^{2} - 4 x + 12 \mathmr{and} y = {x}^{2} - 4 x + 4 + 8$ or

$y = {\left(x - 2\right)}^{2} + 8$, 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 $\left(2 , 8\right)$. y-intercept can be found by

putting $x = 0$ in the equation $y = {x}^{2} - 4 x + 12 \therefore y = 12$

y- intercept is $y = 12$ or at $\left(0 , 12\right)$ . x-intercept can be found by

putting $y = 0$ in the equation $y = {x}^{2} - 4 x + 12$ or

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

Discriminant $D = {b}^{2} - 4 a c = 16 - 48 = - 32 \therefore D < 0$ .

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]