Suppose the curve $y = x^{4} + a x^{3} + b x^{2} + c x + d$ $y=x^4+ax^3+bx^2+cx+d$ has a tangent line when x=0 with equation y=2x+1 and a tangent line when x=1 with equation $y = 2 - 3 x$ $y=2-3x$ , 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.

$y = x^{4} + a x^{3} + b x^{2} + c x + d$ $y = x^4+ax^3+bx^2+cx+d$

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

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 $d=1$

Suppose the curve y=x^4+ax^3+bx^2+cx+dy=x4+ax3+bx2+cx+dy=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=2-3xy=2−3xy=2-3x, how do you find the values of a, b, c, d?