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

the vertex of the function g(x) = f(x 2) .

A quadratic equation's vertex form is defined.

If a quadratic equation is expressed as the following

                          $y=a(x-h)^{2}+k$

It is then said to be in vertex form. Because the vertex point (peak point) occurs on the graph of this equation, it is so named (h,k)

The parabola that the quadratic equation depicts has this point (h,k) as its vertex.

How may the given equation be changed to be in vertex form?

We first remove the x squared coefficient, and then inside the bracket, we strive to create a condition that resembles a perfect square.

So, this is what we have:

                         $f(x)=x^{2}+6 x+3 \\ g(x)=f(x-2)$

Thus, we obtain:

                        $\begin{aligned}&g(x)=f(x-2) \\&g(x)=(x-2)^{2}+6(x-2)+3 \\&g(x)=x^{2}-4 x+4+6 x-12+3 \\&g(x)=x^{2}+2 x-5\end{aligned}$

Vertex form transformation of g(x):

                             $\begin{aligned}&g(x)=x^{2}+2 x-5 \\&g(x)=x^{2}+2 x+1-1-5 \\&g(x)=(x+1)^{2}-9 \\&g(x)=(x-(-1))^{2}-6 \\&g(x)=(x-(-1))^{2}+(-6)\end{aligned}$

By comparing this$a(x-h)^{2}+k$ to, we obtain the coordinates of the vertex of the graph of g(x) as (h,k) = (-1,-6), as shown below.

As a result, the vertex of the function g(x) = f(x 2) .

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