Question

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

Answer 1

Vertex is at `(1,5)` , Y-intercept is at `(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]