



 
Hi Kevin, There is a point $P$ on the curve $y = x^3$ where the tangent to $y = x^3$ at $P$ passes through (0,2). Suppose the xcoordinate of $P$ is $a$ then since $P$ is on the graph of $y = x^3$ the ycoordinate of $P$ is $a^3.$ Use calculus to find the slope of the tangent to $y = x^3$ at $P$ and write the equation of the tangent. This tangent line passes through (0,2). Solve for $a.$ Make sure you verify your answer. Penny  


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