Suppose the curve #y=x^4+ax^3+bx^2+cx+d# has a tangent line when x=0 with equation y=2x+1 and a tangent line when x=1 with equation #y=2-3x#, how do you find the values of a, b, c, d?

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Answer 1 · 2016-10-15T21:23:16

Here's an outline.

#y = x^4+ax^3+bx^2+cx+d#

#y' = 4x^3+3ax^2+2bx+c#

Knowing the equations of the tangent lines at #x=0# and #x = 1# allows us to find #y# and #y'# at those values of #x#.

At #x=0#, we get #y=2(0)+1=1# and #y'=2#

At #x=1#, we get #y=2-3(1)=-1# and #y'=-3#

Using the formulas above we get 4 equations with 4 unknowns;

#x=0# we get

#y = d = 1#

#y' = c = 2#

#x=1# we get

#y = 1+a+b+c+d=-1#

#y' = 4+3a+2b+c = -3#

Solve the system to get

#a=1#, #b=-6#, #c=2#, and #d=1#