# How do you find the polynomial function whose graph passes through (2,4), (3,6), (5,10)?

Simplest solution:

#f(x) = 2x#

General solution:

#f(x) = P(x)(x^3-10x^2+31x-30)+2x#

As stated:

Thus, the following polynomial function is appropriate:

But take note that there are other polynomial functions that also pass through these three points.

Any multiple of a cubic whose zeros are located at those three points can be added, either scalar or polynomial, as follows:

Therefore, the most applicable remedy is:

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Use the points to set up a system of equations. For each point (x, y), you get an equation of the form (f(x) = y).

(f(2) = 4), (f(3) = 6), (f(5) = 10)

Solve this system to find the coefficients of the polynomial function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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