How do you find the vertex and the intercepts for f(x) = 3x^2 - 6x + 8f(x)=3x26x+8?

1 Answer
Mar 28, 2017

Vertex is at (1,5)(1,5) , Y-intercept is at (0,8)(0,8)

Explanation:

f(x)= 3x^2-6x+8 or f(x)= 3(x^2-2x)+ 8 or f(x)= 3(x^2-2x+1) -3+8 or f(x)=3(x-1)^2 -+5 :.

Vertex is at (1,5) [Comparing with standard equation f(x)=a(x-h)^2+k , (h,k) is vertex]

f(x) = 0+0+8=8 ; x=0
Y-intercept is at (0,8) [Obtained by putting x=0 in the equation.]

X-intercept can be obtained by putting y=0 in the equation.
3x^2-6x+8=0 , Discriminant = b^2-4ac = -60 is negative, roots are complex and parabola does not cross x axis

Vertex is at (1,5) , Y-intercept is at (0,8) graph{3x^2-6x+8 [-40, 40, -20, 20]}[Ans]