Respuesta :

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If [tex]f(x)=x^2-6x+3[/tex], then [tex]f(x-2)=(x-2)^2-6(x-2)+3[/tex].

Let's simplify that.

Distribute with [tex]-6(x-2)[/tex]:

[tex]f(x-2)=(x-2)^2-6x+12+3[/tex]

Combine the end like terms [tex]12+3[/tex]:

[tex]f(x-2)=(x-2)^2-6x+15[/tex]

Use [tex](x-b)^2=x^2-2bx+b^2[/tex] identity for [tex](x-2)^2[/tex]:

[tex]f(x-2)=x^2-4x+4-6x+15[/tex]

Combine like terms [tex]-4x-6x[/tex] and [tex]4+15[/tex]:

[tex]f(x-2)=x^2-10x+19[/tex]

We are given [tex]g(x)=f(x-2)[/tex].

So we have that [tex]g(x)=x^2-10x+19[/tex].

The vertex happens at [tex]x=\frac{-b}{2a}[/tex].

Compare [tex]x^2-10x+19[/tex] to [tex]ax^2+bx+c[/tex] to determine [tex]a,b,\text{ and } c[/tex].

[tex]a=1[/tex]

[tex]b=-10[/tex]

[tex]c=19[/tex]

Let's plug it in.

[tex]\frac{-b}{2a}[/tex]

[tex]\frac{-(-10)}{2(1)}[/tex]

[tex]\frac{10}{2}[/tex]

[tex]5[/tex]

So the [tex]x-[/tex] coordinate is 5.

Let's find the corresponding [tex]y-[/tex] coordinate by evaluating our expression named [tex]g[/tex] at [tex]x=5[/tex]:

[tex]5^2-10(5)+19[/tex]

[tex]25-50+19[/tex]

[tex]-25+19[/tex]

[tex]-6[/tex]

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is [tex]a(x-h)^2+k[/tex] where the vertex is [tex](h,k)[/tex].

Let's put [tex]f[/tex] into this form.

We are given [tex]f(x)=x^2-6x+3[/tex].

We will need to complete the square.

I like to use the identity [tex]x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2[/tex].

So If you add something in, you will have to take it out (and vice versa).

[tex]x^2-6x+3[/tex]

[tex]x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2[/tex]

[tex](x+\frac{-6}{2})^2+3-3^2[/tex]

[tex](x+-3)^2+3-9[/tex]

[tex](x-3)^2+-6[/tex]

So we have in vertex form [tex]f[/tex] is:

[tex]f(x)=(x-3)^2+-6[/tex].

The vertex is (3,-6).

So if we are dealing with the function [tex]g(x)=f(x-2)[/tex].

This means we are going to move the vertex of [tex]f[/tex] right 2 units to figure out the vertex of [tex]g[/tex] which puts us at (3+2,-6)=(5,-6).

The [tex]y-[/tex] coordinate was not effected here because we were only moving horizontally not up/down.