Question

The function f(x)=x^2 -6x+3 is transformed such that g(x)=f(x-2). Find the vertex of g(x).

(5,6)

(5,-8)

(1,-6)

(3,-8)

Answer 1

`g(x) = (x-2)^2 -6(x-2) +3`

`g(x) = x^2 -4x +4 -6x +12 +3`

`g(x) = x^2 -10 x +19`

`g(x) = x^2 -10 x + 25 +19 -25`

`g(x) = (x-5)^2 -6`

`y= (x-h)^2 -kk`

Where h,k represent the vertex and we got:

`(h,k)= (5,6)`

(5,6)

Explanation:

We have this original function given :

`f(x) = x^2 -6x +3`

And we want to find the vertex for this new function
`g(x) = f(x-2)` and we have:

`g(x) = (x-2)^2 -6(x-2) +3`

And solving the square we got:

`g(x) = x^2 -4x +4 -6x +12 +3`

And adding similar terms we got:

`g(x) = x^2 -10 x +19`

Now we can complete the square like this:

`g(x) = x^2 -10 x + 25 +19 -25`

`g(x) = (x-5)^2 -6`

The general equation is given by:

`y= (x-h)^2 -kk`

Where h,k represent the vertex and we got:

`(h,k)= (5,6)`

(5,6)