F (x)=x^2-2х
f’= (f(x+h) - f(x))/h

If you're asking about the difference quotient alone, you have

f (x) = x ² - 2x

which means

f (x + h) = (x + h)² - 2 (x + h)

so that

(f (x + h) - f (x)) / h = (((x + h)² - 2 (x + h)) - (x ² - 2x)) / h

… = (x ² + 2xh + h ² - 2x - 2h - x ² + 2x) / h

… = (2xh + h ² - 2h) / h

*and you would stop here*.

On the other hand, if you're talking about the derivative (which I believe you are, since your question says f '), take the limit as h approaches 0. In particular, this means h ≠ 0, so there is some cancellation:

(f (x + h) - f (x)) / h = (2xh + h ² - 2h) / h

… = 2x + h - 2

Then as h approaches 0, you're left with

f ' (x) = 2x - 2

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source