Question

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

Answer 1

Answer:

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

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

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

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

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

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

Where h,k represent the vertex and we got:

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

(5,6)

Step-by-step 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)