Grade 11 math question

[simonehope]

For the function g(x)=sqrt x2-6x+9 -3,(-3 not under sqrt) determine:
b) g(d)

sqrt (d)^2-6(d)+9-3
sqrt (d+3)(d-3)-3
(d-3)-3
g(d)=d+9

 

[JeffM]

Please give the complete and exact question as we request in our guidelines.

[MATH]g(x) = \sqrt{x^2 - 6x + 9} - 3[/MATH].

We also ask in our guidelines that you either show what work you were able to do (even if you know it is wrong) or else tell us briefly just why you cannot even start.

 

[HallsofIvy]

if g(d)

If g(d) is what??

what is g(x)= sqrt x^2-6x+9 -3
(-3 not under sqrt)

Please use parenthesese! I think you mean g(x)=sqrt(x^2- 6x+ 9)- 3
Do you know that (x- 3)^2= x^2- 6x+ 9?..................................................................corrected

 

[simonehope]

If g(d) is what??


Please use parenthesese! I think you mean g(x)=sqrt(x^2- 6x+ 9)- 3
Do you know that (x- 3)^2= x^3- 6x+ 9?

No I don't know (x- 3)^2= x^3- 6x+ 9. Can you explain?

 

[simonehope]

x2

*g(x)=sqrt x^2-6x+9 -3,(-3 not under sqrt)

 

[Deleted member 4993]

No I don't know (x- 3)^2= x^3- 6x+ 9. Can you explain?

Can you multiply the following:

(x - 3) * (x - 3) = ?

 

[simonehope]

x^2-3x-3x+9?
I don't understand where the (x-3)^2= x^3 comes from though.

 

[JeffM]

WHAT IS THE QUESTION you were asked? g(d) means nothing.

 

[simonehope]

WHAT IS THE QUESTION you were asked? g(d) means nothing.

this is the question I was asked:
For the function g(x)=sqrt x^2-6x+9 -3,(-3 not under sqrt) determine:
b) g(d).
that's it.

g(x) is the same as f(x).
d is what you are subbing in for x.

 

[Deleted member 4993]

(x- 3)^2= x^3- 6x+ 9?

That was a typo.

It should have been:

(x- 3)^2= x^2- 6x + 9?

 

[simonehope]

That was a typo.

It should have been:

(x- 3)^2= x^2- 6x + 9?

where does (x-3)^2 come from?
I understand if you where to expand x^2-6x into (x+something)(x-something) but I don't understand what (x-3)^2 is.

 

[JeffM]

this is the question I was asked:
For the function g(x)=sqrt x^2-6x+9 -3,(-3 not under sqrt) determine:
b) g(d).
that's it.

g(x) is the same as f(x).
d is what you are subbing in for x.

Well then step one is to do the substitution

[MATH]\sqrt{d^2 - 6d + 9} - 3 = \sqrt{(d - 3)^2} - 3 = WHAT? [/MATH]

 

[simonehope]

Well then step one is to do the substitution

[MATH]\sqrt{d^2 - 6d + 9} - 3 = \sqrt{(d - 3)^2} - 3 = WHAT? [/MATH]

where is \sqrt{(d - 3)^2} - 3 coming from?

 

[Deleted member 4993]

Can you multiply the following:

(x - 3) * (x - 3) = ?

and you said:

x^2-3x-3x+9

x^2-3x-3x+9 = x^2 - 6x + 9

replace 'x' by 'd' and you get:

(d - 3)^2 = d^2 - 6d + 9

 

[JeffM]

where is \sqrt{(d - 3)^2} - 3 coming from?

[MATH](d - 3)^2 = (d - 3)(d - 3) = d(d - 3) - 3(d - 3) = d^2 - 3d - 3d + 9 = d^2 - 6d + 9.[/MATH]
[MATH]\sqrt{d^2 - 6d + 9} = \sqrt{(d - 3)^2} = d - 3.[/MATH]
When you have something inside a square root, it is always a good idea to look ro see whether it is a square of something so you can simplify.

 

[simonehope]

and you said:



x^2-3x-3x+9 = x^2 - 6x + 9

replace 'x' by 'd' and you get:

(d - 3)^2 = d^2 - 6d + 9

okay. then what?

 

[simonehope]

[MATH](d - 3)^2 = (d - 3)(d - 3) = d(d - 3) - 3(d - 3) = d^2 - 3d - 3d + 9 = d^2 - 6d + 9.[/MATH]
[MATH]\sqrt{d^2 - 6d + 9} = \sqrt{(d - 3)^2} = d - 3.[/MATH]
When you have something inside a square root, it is always a good idea to look ro see whether it is a square of something so you can simplify.

where did +9 go?

 

[JeffM]

where did +9 go?

(-3)^2 = + 9, no?

 

[simonehope]

g(x) = sqrt{x^2 - 6x + 9} - 3.

This is the original formula. I don't know what happened to the +9 -3

 

[pka]

g(x) = sqrt{x^2 - 6x + 9} - 3.
This is the original formula. I don't know what happened to the +9 -3

You need to know that \(\sqrt{x^2}=|x|\) so that \(\sqrt{x^2-6x+9}-3=\sqrt{(x-3)^2}-3=|x-3|-3\)