How do you find the equation of the line #y = x^3 - 5x^2 + 2x +1# tangent to the curve and its point of inflection?

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Answer 1 · 2017-08-16T06:03:22

The equation of the line is #y=-19/3x+152/27#

Let #f(x)=x^3-5x^2+2x+1#

Taking the first derivative

#f'(x)=3x^2-10x+2#

Taking the second derivative

#f''(x)=6x-10#

The point of inflection is when #f''(x)=0#

#6x-10=0#, #=>#, #x=10/6=5/3#

The slope of the curve when #x=5/3# is

#f'(5/3)=3*(5/3)^2-10*5/3+2=25/3-50/3+2=-25/3+2=-19/6#

When #x=5/3#, #=>#

#f(5/3)=(5/3)^3-5*(5/3)^2+10/3+1=125/27-125/9+10/3+1#

#=(125-375+90+27)/27=-133/27#

The eqaution of the tangent at the point #=(5/3, -133/27)# is

#y-f(5/3)=f'(5/3)(x-5/3)#

#y+133/27=-19/3(x-5/3)#

#y=-19/3x+95/9-133/27=-19/3x+152/27#

graph{(y-x^3+5x^2-2x-1)(y+19/3x-152/27)=0 [-18.56, 21.99, -14, 6.28]}