Math help (algebraic functions)

[Nanobyte]

OK, we're reviewing right now in AP Cal, and I don't understand how this one function works. Here goes:

f(x)= x²-x+3
f(a+h)= a²-a+2ah+h²-h+3

What I don't understand is how the 2ah comes into all this. They cover functions in the book, but not f(x+z) only f(x)+g(x) or (f+g)(x). Can someone help?

[Vaskez]

In words: f(x) is a function acting on the unknown x (representing any unknown value) that transforms this unknown value into itself squared - itself + 3.

note: a^2 means "a squared" but I can't do superscript here

So all you have to do to apply this function to any other value is substitute this other value in the place of x.
So if we want to work out f(a+h) simply substitute (a+h) everywhere where there is x in the definition of f(x)

f(x) = x^2 - x + 3
so
f(a+h) = (a+h)^2 - (a+h) + 3
=> (a+h)*(a+h) - (a+h) + 3 (multiply out)
=> a^2 + ah + ha + h^2 -a -h + 3
=> a^2 + 2ah + h^2 -a -h +3

and that's where the 2ah comes from, simple as that

[ 08-12-2003, 07:58 PM: Message edited by: Vaskez ]

[Nanobyte]

Wow, I can't believe I didn't get that. Thanks Vaskez [img]smile.gif[/img]

BTW, sub for subscript and sup for superscript.

[InsaneBane]

Originally posted by Vaskez: f(x) = x^2 - x + 3 so f(a+h) = (a+h)^2 - (a+h) + 3 => (a+h)*(a+h) - (a+h) + 3 (multiply out) => a^2 + ah + ha + h^2 -a -h + 3 => a^2 + 2ah + h^2 -a -h +3

or more precisely written:

f(x) = x^2 - x + 3 <=>
f(a+h) = (a+h)^2 - (a+h) + 3 <=>
f(a+h) = (a+h)*(a+h) - (a+h) + 3 <=>
f(a+h) = a^2 + ah + ha + h^2 -a -h + 3 <=>
f(a+h) = a^2 + 2ah + h^2 - a - h + 3

If you want to pass it on to your math teacher [img]smile.gif[/img]

[img]graemlins/greenbounce.gif[/img]
Insane

[Bruce The Aussie]

Originally posted by Vaskez: In words: f(x) is a function acting on the unknown x (representing any unknown value) that transforms this unknown value into itself squared - itself + 3. note: a2 means "a squared" but I can't do superscript here So all you have to do to apply this function to any other value is substitute this other value in the place of x. So if we want to work out f(a+h) simply substitute (a+h) everywhere where there is x in the definition of f(x) f(x) = x2 - x + 3 so f(a+h) = (a+h)2 - (a+h) + 3 => (a+h)*(a+h) - (a+h) + 3 (multiply out) => a2 + ah + ha + h2 -a -h + 3 => a2 + 2ah + h2 -a -h +3 and that's where the 2ah comes from, simple as that

hello^hello

just quote this and you can find out how to do superscript [img]tongue.gif[/img]

cool...i've been trying to understand this for my Pure 1 exams this november....i don't like it much.....

[ 08-13-2003, 03:19 AM: Message edited by: Bruce The Aussie ]