Question

Will mark brainliest!!!plz helppp

Answer 1

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If $f(x)=x^2-6x+3$, then $f(x-2)=(x-2)^2-6(x-2)+3$.

Let's simplify that.

Distribute with $-6(x-2)$:

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

Combine the end like terms $12+3$:

$f(x-2)=(x-2)^2-6x+15$

Use $(x-b)^2=x^2-2bx+b^2$ identity for $(x-2)^2$:

$f(x-2)=x^2-4x+4-6x+15$

Combine like terms $-4x-6x$ and $4+15$:

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

We are given $g(x)=f(x-2)$.

So we have that $g(x)=x^2-10x+19$.

The vertex happens at $x=\frac{-b}{2a}$.

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

$a=1$

$b=-10$

$c=19$

Let's plug it in.

$\frac{-b}{2a}$

$\frac{-(-10)}{2(1)}$

$\frac{10}{2}$

$5$

So the $x-$ coordinate is 5.

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

$5^2-10(5)+19$

$25-50+19$

$-25+19$

$-6$

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

ANOTHER WAY:

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

Let's put $f$ into this form.

We are given $f(x)=x^2-6x+3$.

We will need to complete the square.

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

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

$x^2-6x+3$

$x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2$

$(x+\frac{-6}{2})^2+3-3^2$

$(x+-3)^2+3-9$

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

So we have in vertex form $f$ is:

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

The vertex is (3,-6).

So if we are dealing with the function $g(x)=f(x-2)$.

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

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