Tangent line of parabola?

Question

If f(x)=x^3-5x+1, find f'(1)

 

f'(1)= ???

 

Use it to find an equation of the tangent line to the parabola y=x^3-5x+1 at the point (1,-3).

 

y=???

Answer 1

First, Maria, your 3 exponent could have been 2 for f(x) to be properly called a parabola !

 

A. Assuming you made a typo ( a mistake) f(x) = x² - 5x + 1

    f(1) = 1² - 5(1) + 1 = -3

    The slope of the tangent line is y - (-3) = f’(1)[x - (1)]

    where f’(1) is the value of the derivative of f taken at x = 1

    f’(x)=2x - 5 ------> f’(1)=2(1) - 5 = -3

 

    y + 3 = -3(x - 1) -----> y = -3x

 

B. Assuming you call improperly f(x) = x³ - 5x + 1 a parabola,

    f(1) = 1³ - 5(1) + 1 = -3

    Slope f’(x) = 3x² - 5 ------> f’(1)=3(1²) - 5 = -2

 

    y + 3 = -2(x - 1) -----> y = -2x + 2 - 3

                                -----> y = -2x - 1