Question

The function f(x) = x2 + 6x + 3 is transformed such that g(x) = f(x − 2). Find the vertex of g(x).

Answer 1

Answer:

(-1,-6)

Step-by-step explanation:

$f(x)=x^2+6x+3$

$g(x)=f(x-2)$

$g(x)=(x-2)^2+6(x-2)+3$  (Replaced x in f with (x-2))

$g(x)=x^2-4x+4+6x-12+3$ (Used $(x+b)^2=x^2+2bx+b^2$ and distributive property)

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

$g(x)=x^2+2x-5$  (Simplified)

The vertex of a parabola occurs at $(\frac{-b}{2a},g(\frac{-b}{2a})$.

Let's find $\frac{-b}{2a}$ first.

$\frac{-2}{2(1)}=-1$

Now we can obtain $g(\frac{-b}{2a})$ which is $g(-1)$ in this case:

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

$g(-1)=(-1)^2+2(-1)-5$

$g(-1)=1-2-5$

$g(-1)=-1-5$

$g(-1)=-6$.

The vertex is (-1,-6).