suppose the curve y=x^4+ax^3+bx^2+cx+d has a tangent line when x=o with the equation y=2x+1 and a tangent line when x=1 with the equation y=2-3x. find the values of a, b, c, d
suppose the curve y=x^4+ax^3+bx^2+cx+d has a tangent line when x=o with the equation y=2x+1 and a tangent line when x=1 with the equation y=2-3x. find the values of a, b, c, d
Well,
$\displaystyle \frac{dy}{dx} = 4x^3 + 3ax^2 + 2bx + c$
We know that when x = 0 we have a tangent line with a slope of 2. So:
$\displaystyle 2 = c$
Similarly we know that when x = 1 we have a tangent line with a slope of -3. So:
$\displaystyle -3 = 4 + 3a + 2b + c$
You might think we are done, but oh no! There's more:
We also know that at the point x = 0 the tangent line shares a point with the function y, so y = 2x + 1 at x = 0 gives the point (0, 1). Thus
$\displaystyle y=x^4+ax^3+bx^2+cx+d \implies 1 = d$
Similarly the tangent line shares a point with the function when x = 1, so y = -3x + 2 gives the point (1, -1):
$\displaystyle y=x^4+ax^3+bx^2+cx+d \implies 1 = 1 -a + b -c + d$
You should be able to solve this from here.
-Dan