Question

Find the difference quotient for the equation f(x)=-x^2+x ?

Answer 1

Do you have to show work, or can you take a shortcut?

Answer 2

show work if possible

Answer 3

Ok, you know what 'difference quotient' means right?
It's just like the slope formula for straight lines, but uses differentials.
\[DQ=\frac{f(x+\Delta x)-f(x)}{\Delta x}\]

Answer 4

Do you know how to set it up from there?

Answer 5

no

Answer 6

(-(x+h)^2+(x+h))-(-x^2+x))/h = (-(x^2+2xh+h^2)+x+h+x^2-x)/h =
(-x^2-2xh-h^2+x+h=x^2-x)/h = (-2xh-h^2+h)/h = -2x+1

Answer 7

which is the derviative of -x^2 + x

Answer 8

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

Answer 9

hope that helps

Answer 10

the first equal sign on the second line should be +

Answer 11

(-(x+h)^2+(x+h))-(-x^2+x))/h = (-(x^2+2xh+h^2)+x+h+x^2-x)/h =
(-x^2-2xh-h^2+x+h+x^2-x)/h = (-2xh-h^2+h)/h = -2x+1

Answer 12

have fun in calculus, its really fun:)

Answer 13

does this make sense @jalotaibi ?

Answer 14

If \[f(x)=-x^2+x\] then the difference quotient,
\[\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{-(x+\Delta x)^2+(x+\Delta x)-(-x^2+x)}{\Delta x}.\]

Answer 15

Then just simplify from there.