What is the equation of the tangent line of #f(x)=(5+4x)^2 # at #x=7#?

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Answer 1 · 2016-11-23T18:04:49

The equation is #y=264x-759#

To calculate the tangent, we need the gradient at #x_0=7#

#f(x_0)=33^2=1089#

#f'(x)=2*(5+4x)*4=8(5+4x)#

#f'(x_0)=8*33=264#

The equation of the tangent is

#y-y_0=f'(x_0)(x-x_0)#

#y-1089=264(x-7)#

#y=264x-1848+1089#

#y=264x-759#

graph{(y-(5+4x)^2)(y-264x+759)=0 [-299.3, 309.2, -83.8, 220.8]}