What is the equation of the normal line of #f(x)=5x^3+2x^2+x-2# at #x=2#?

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Answer 1 · 2016-09-18T03:07:59

#x+69y=3314#

#f(2)=40+8+2-2=48#

The tangent line is: #y=kx+n#, where #k=f'(x)#. So,

#f'(x)=15x^2+4x+1#

#f'(2)=69 => k=69#

The normal line is: #y=k_1x+n_1#, where #k_1=-1/k#.

#k_1 = -1/69#

#48 = -1/69*2+n_1 => n_1 = 48+2/69 = 3314/69#

Finally,

#y=-1/69x+3314/69# or

#x+69y=3314#