Equation of tangent line to the graph f(x)= (x-2)(x^2-3x-1) at the point (-1,-9)
is the equation y+9=9(x+1)
or am i missing a step

Question

Equation of tangent line to the graph f(x)= (x-2)(x^2-3x-1) at the point (-1,-9)
is the equation y+9=9(x+1) 
or am i missing a step

amistre64 · Answer

whats your f' ?

amistre64 · Answer

f(x)= (x-2)(x^2-3x-1)

f'(x)= (x-2)'(x^2-3x-1) + (x-2)(x^2-3x-1)'
      = (1)(x^2-3x-1) + (x-2)(2x-3)

f'(-1) =  ((-1)^2-3(-1)-1) + ((-1)-2)(2(-1)-3)
         =  (1+3-1) + (-3)(-5)
         =  3 + 15
         =  18

or did i miss the math?

amistre64 · Answer

the rest looks good; change the slope tho

liizzyliizz · Answer

oh ok that must of been my math then. i see where i went wrong i just realized i missed one term when using the product rule. but thank you !

amistre64 · Answer

;)  youre welcome