Question 1191707: The tangent to the curve y = ax^3 + bx at the point (1,3) crosses the y-axis at (0,-4).Find the value of a and of b.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The slope of the tangent line is 3ax^2+b, and at point (1,3) it is 3a+b
We know two points of the tangent line, and their slope is -7/-1 or 7
so 3a+b=7
Additionally a+b=3 since the curve goes through (1,3)
Subtract the second from the first to get 2a=4, a=2, so b=1, the answer.
the curve is 2x^3+x. The tangent line is y-y1=m(x-x1) m slope (x1, y1) point
so it is y-3=7(x-1) or y-7x-4
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