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I have a cubic function:

$ y = 3 + 5x + x^2 - x^3$

and am supposed to prove that it goes through the point C with coordinates (3, 0). The question also asked to find 2 of its stationary points which I was able to find by differentiating the cubic equation and setting it to zero.

But I'm not sure how to prove that this graph goes through (3, 0).

These are the only two methods I could think of: 1- Just plug in 3 and 0 and see that it works. 2- Divide the equation by $(x-3)$ and show that there's no remainder.

This is an A'Levels calculus question. What's the right approach to follow here?

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2 Answers 2

up vote 3 down vote accepted

Option 1 is the easier one. "Plug in" $3$ for $x$. Is the result $0$?

In general, if we have the curve $y=f(x)$, to find the $y$-coordinate at the point with $x$-coordinate $a$, we just calculate $f(a)$.

Dividing your polynomial by $x-3$ will work, but it is quite a bit more painful. And the procedure will not work when $f(x)$ is not a polynomial.

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Like the answer below me said plug in 3. Notice f(x) = 3 + 15 + 9 - 27 = 0. So we know x-3 is a factor of this solution. You could also see this would have no remainder,

If we apply synthetic division (i dont know how to properly do it right here but ill give it a shot)

3 [ -1 1 5 3], if we apply synthetic we get -1 -2 -1 0 which just boils down to

-x^2 - 2x-1 with remainder 0. So yes (3,0) go through that function by the remainder theorem so we factored it down to

-(x-3)(x^2+2x+1) = -(x-3)(x+1)^2 so we can also see -1 is a root :)

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