Let: f(x) = 5x^3 + 13x - 3 The derivative function is: f ' (x) = ((5(x+h)^2 + 13(x+h) - 3) - (5x^2 + 13x - 3)) / h

The f ' (x) = df/dx = 10x + 13

Based on this information, I need to find the:: tangent line for f(x) where x = 2.
Can someone please help me find this? I need immediate help please.

I think the equation would be y = 33(2) + b, but I'm not sure if this equation is right, and if this equation is right, I don't know what b would be.

Please help me find the tangent line!

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

the answer is 1 becuz if it were 1.0 that would rip apart the entire so ya. : )

~potato.

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