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

May 19, 2018

Simplest solution:

$f \left(x\right) = 2 x$

General solution:

$f \left(x\right) = P \left(x\right) \left({x}^{3} - 10 {x}^{2} + 31 x - 30\right) + 2 x$

#### Explanation:

Given:

$\left(2 , 4\right)$, $\left(3 , 6\right)$, $\left(5 , 10\right)$

Note that each $y$ coordinate is twice the corresponding $x$ coordinate.

So a suitable polynomial function is:

$f \left(x\right) = 2 x$

Note however that this is not the only polynomial function passing through these three points.

We can add any multiple (scalar or polynomial) of a cubic whose zeros lie at those three points, namely:

$\left(x - 2\right) \left(x - 3\right) \left(x - 5\right) = {x}^{3} - 10 {x}^{2} + 31 x - 30$

Hence the most general solution is:

$f \left(x\right) = P \left(x\right) \left({x}^{3} - 10 {x}^{2} + 31 x - 30\right) + 2 x$

for any polynomial $P \left(x\right)$.