Question 202334
Find the inverse of quadratic function, graph function and its inverse in the same coordinate plane.
y=x^2-2x+1 
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Interchange x and y to get:
x = y^2 - 2y + 1
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Solve for "y":
y^2 - 2y + (1-x) = 0
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Use the quadratic formula to solve for "y":
y = [2 +- sqrt(4 - 4(1-x)]/2
y = [2 +- sqrt(4(1-(1-x)[/2
y = [2 + 2sqrt(x)]/2 or y = [2 - 2sqrt(x)]/2
y = [1 + sqrt(x)] or y = [1-sqrt(x)]
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Graph:
{{{graph(400,300,-10,10,-10,10,x^2-2x+1,(1+(x)^(0.5)),(1-(x)^(0.5)))}}}
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Cheers,
Stan H.