Please I need help with this algebra problem
With the 3 equations
f1(x)=0
f2(x)=1/8(x+16)^2-98
f3(x)=2x+24
Determine an equation for this function with the 3 pieces above.(Refer to creating the polynomial functions)
I do not know how does everyone get this answer g(x) = f2(x)-f3(x)
= (1/8(x+16)^2-98)-(2x+24)
= 1/8 x^2 + 2x - 90
After seeing how you responded to Steve when he helped you with this question ....
http://www.jiskha.com/display.cgi?id=1479505703
The g(x) I suggested was a last-ditch effort to come up with some function using f1,f2,f3. You repeatedly refused to specify what you meant when you said
Determine an equation for this function
"this function" indicates that there is some function you want to create using f1,f2,f3. But you never say what it is; I made one up, just to illustrate how to use them. If you have some other function, we'll be happy to help with its derivation.
Just as a final effort to be helpful, despite your intransigence,
(1/8(x+16)^2-98)-(2x+24)
= 1/8 (x^2+32x+256) - 98 - 2x - 24
= 1/8 x^2 + 4x + 32 - 98 - 2x - 24
= 1/8 x^2 + 2x + 90
You really need to do some of the work on your own, rather than just sulking when the entire solution is not provided.
Did Steve's additional explanation help you?
To determine an equation for the function using the given pieces, we need to subtract the third equation from the second equation. This will give us the equation for the function that combines f2(x) and f3(x).
Let's go through step by step:
1. Start with f2(x) = 1/8(x+16)^2 - 98 and f3(x) = 2x + 24.
2. Subtract f3(x) from f2(x): f2(x) - f3(x) = (1/8(x+16)^2 - 98) - (2x + 24).
3. Distribute the negative sign across f3(x): f2(x) - f3(x) = 1/8(x+16)^2 - 98 - 2x - 24.
4. Simplify the expression by combining like terms: f2(x) - f3(x) = 1/8(x+16)^2 - 2x - 122.
Therefore, the equation for the function g(x) is g(x) = f2(x) - f3(x) = 1/8(x+16)^2 - 2x - 122.
Note: The expression 1/8 x^2 + 2x - 90 you mentioned is incorrect because you have omitted the square term and made a mistake in the constant term.