How do you find the inverse of $y = 2 x^{2} - 3 x + 1$ $y = 2x^2 - 3x +1$ and is it a function?

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Answer 1 · 2016-06-05T20:48:57

To find the inverse you need to write $x =$ $x=$ something, then it is exactly the same way as to solve the second order equation:

$y = 2 x^{2} - 3 x + 1$ $y=2x^2-3x+1$

$2 x^{2} - 3 x + 1 - y = 0$ $2x^2-3x+1-y=0$

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot (1 - y)}}{4}$ $x=(3\pm sqrt(9-4*2*(1-y)))/4$

$x = \frac{3 \pm \sqrt{1 - 8 y}}{4}$ $x=(3\pm sqrt(1-8y))/4$ .

This is the inverse. It is not a function because for one $y$ $y$ you have two $x$ $x$ , one with the plus solution and one with the minus.

How do you find the inverse of y=2x2−3x+1y = 2x^2 - 3x +1 and is it a function?