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Find the slope of a line $L$ that tangent to the graph of $y = x^3$ and passes through the point $(0,2000)$.

Well, I am new to this concept, to me, slope means $\dfrac{dy}{dx}$, but I get $\dfrac{dy}{dx}=3x^2$, but I do not understand what I need to do after this. Please help me. Thank you.

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Since $\dfrac{dy}{dx} = 3x^2$, the line tangent to $y = x^3$ at the point $(a,a^3)$ will have a slope of $3a^2$.

If you use the point-slope formula, the equation of the line is $y-a^3 = 3a^2(x-a)$.

Now, for what value of $a$ does this line pass through $(x,y) = (0,2000)$?

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  • $\begingroup$ Thanks! that just did the work. You helped a lot! $\endgroup$ – Swadhin May 15 '15 at 6:23
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Hint: You could find the line joining the points $(x,x^3)$ and $(0,2000)$, which will depend on $x$, and then determine which of these lines has slope $3x^2$

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