Work out the value of k in $y = \frac13 x^3 - x^2 -3x + k$

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I'm finding this question really difficult, it's on the further maths GCSE and has a cubic graph picture below it. If that would help, let me know.

$$y = \tfrac13 x^3 - x^2 -3x + k$$

where $k$ is a constant.

The $x$ axis is a tangent to the curve at its minimum point. Work out the value of $k$.

I know basic differentiation but don't know where to start on this. I'm a GCSE Further Maths student so I don't have a vast amount of mathematical experience; please be patient and explain in steps.

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I assume when you say minimum point, you mean local minimum.

To find the local maximum or minimum of a function, you first find the derivative. In this case, the derivative is $x^2-2x-3$. Next you would set the derivative equal to $0$, which would give you $x=3$ and $x=-1$.

Since this is a cubic function, one of these is a local max and the other is a local min. To find which is a local min, you would first find the 2nd derivative of the function (the derivative of the derivative) which is $2x-2$. The $x$ value for which the 2nd derivative is positive is the local min. Therefore in this case, $x=3$ is the local min.

Since the $x$-axis is tangent to the local min, that means when $x=3$, $y$ must equal $0$. You can now just plug those values into the equation and solve for $k$.