College algebra funtions f(x+h)-f(x) over h?

Question

So this week we had a sub in our collage math class and I'm realy confused. I don't want to fall behind cause I've gotten straite A's so I was wondering if any of you could double check my work and if I did anything roung can u point it out. If you answer well ill give you 5 stars best answer.

So here it is
Considering the following f(x)=-2x^2-x+1 find f(x+h)-f(x) all over h

So here is my work

f(x+h)-f(x) over h= -2(x+h)^2-(x+h)+1-(-2x^2-x+1) all over h

So then I worked it out to
=-2x^2 -2h^2-x-h+1+2x^2+x-1 over h

Witch then simplifyed to
-2h^2-h over h

Answer 1

You did this part wrong!

(x+h)^2 = x^2 + 2hx + h^2

You missed the middle term.



Overall, because of that, you missed -4hx

Answer 2

starting here, where you put



f(x+h)-f(x) over h= -2(x+h)^2-(x+h)+1-(-2x^2-x+1) all over h



I think you made an error on the next step, first do the exponent (x+h)^2 = x^2 + 2xh + h^2 then put that back into the equation.



f(x+h)-f(x) over h= -2(x^2 + 2xh + h^2) -(x + h) +1 - (-2x^2 - x + 1) all over h



Now distribute to get rid of all the parenthesis, so the first "-2" times everything in that first set of ( ) and for the other ( ) you should notice they have a "-" before it, which just will change the terms inside the ( ) to its opposite.



= -2x^2 - 4xh - 2h^2 - x - h + 1 + 2x^2 + x - 1 all over h. Now combine like terms



= -2h^2 - 4xh - h all over h (notice (-2x^2 + 2x^2 = 0, same with -x + x, 1 -1 )



You should see the 3 terms in the numerator have an h, so cancel out an h in each term, only 1 though, and you can get rid of the h in the denominator then.



= -2h - 4x - 1

Answer 3

Close.



f(x+h) = -2(x+h)² - (x+h) + 1 = -2x² - 2h² - 4xh - x - h + 1

f(x) = -2x² - x + 1

The difference is -2h² - 4xh - h

Divide that by h, and you get -2h -4x - 1



Next week's lesson: As h approaches 0, the answer approaches -4x-1. That's the Derivative of this F function.

Answer 4

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Answer 5

f(x) = -2x² - x + 1



[f(x + h) - f(x)] / h

[ (-2(x + h)² - (x + h) + 1) - (-2x² - x + 1) ] / h

[ (-2(x² + 2xh + h²) - x - h + 1) + 2x² + x - 1 ] / h

[ ((-2x² - 4xh - 2h²) - x - h + 1) + 2x² + x - 1 ] / h

[ -2x² - 4xh - 2h² - x - h + 1 + 2x² + x - 1 ] / h

[ -4xh - 2h² - h ] / h

h (-4x - 2h - 1) / h

-4x - 2h - 1

Answer 6

(x+h)^2 = x^2 + 2hx + h^2



Try it again with the above correction



p.s. [f(x+h)-f(x)]/h is called the 'difference quotient'. It will be something you learn more about in calculus (if you take it).

Answer 7

f(x+h)-f(x) over h= -2(x+h)^2-(x+h)+1-(-2x^2-x+1) all over h

f(x+h)-f(x) over h= -2(x^2 + 2hx +h^2)-(x+h)+1-(-2x^2-x+1) all over h

f(x+h)-f(x) over h= -2x^2 - 4hx - 4h^2 -x - h +1 + 2x^2 + x - 1 all over h





Collect 'like' terms

f(x+h)-f(x) over h= (- 4hx - 4h^2 - h) / h



f(x+h)-f(x) over h= (- 4x - 4h - 1) Done!!!!