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

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Answer 1 · 2018-05-19T07:11:38

Simplest solution:

$f (x) = 2 x$ $f(x) = 2x$

General solution:

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

Given:

$(2, 4)$ $(2, 4)$ , $(3, 6)$ $(3, 6)$ , $(5, 10)$ $(5, 10)$

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

So a suitable polynomial function is:

$f (x) = 2 x$ $f(x) = 2x$

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:

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

Hence the most general solution is:

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

for any polynomial $P (x)$ $P(x)$ .