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

1 Answer
Jun 11, 2017

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

Explanation:

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]