Question

Please help 50 points

Answer 1

Answer:

x-3

Step-by-step explanation:

$\sqrt{x^2 -x+9$

First rewrite 9 as 3^2 so... $\sqrt{x^2 -6x+3^2}$

Next multiply the equation by 2ab so... 2ab =  2 · x · -3

Simplify so... 2ab = -6x

Then use the Perfect Square Trinomial rule: a^2 - 2ab + b^2 = (a-b)^2

so.. a = x and b = -3 to look like this: $\sqrt(x- 3^2)$

And then finally pull out terms under the radical to get...

x-3

Answer 2

Answer:

0 < y <= 6

Step-by-step explanation:

Given:

√(x² - 6x + 9)

if -3 <= x < 3

you can write (x² - 6x + 9)

as (x-3) * (x- 3) which is (x-3)²

so now you have √{(x-3)²}, which is the same as the absolute value of x-3. In mathematics it is written as:

| x - 3 |

So √(x² - 6x + 9) = | x - 3 |

See the graph in the attachment.

for -3 <= x < 3

you get the y values bigger then ( but not equal to) 0 and smaller then or equal to 6.

This is written as 0 < y <= 6