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# Range of a Polynomial (Tomato subjective 66)

problem: If c is a real number with 0 < c < 1, then show that the values taken by the function y = $${frac {x^2+2x+c}{x^2+4x+3c}}$$ , as x varies over real numbers, range over all real numbers.

solution:       y = $${\frac {x^2+2x+c}{x^2+4x+3c}}$$
or $${yx^2}$$ + 4yx + 3cy = $${x^2}$$ + 2x + c
or $${x^2}$$ (y-1) + (4y-2)x + 3cy – c = 0

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June 15, 2015
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