# SOLUTION: Find the inverse of quadratic function, graph function and its inverse in the same coordinate plane. y=x^2-2x+1 How do I find the answer?

Inverse of a quadratic function with domain restriction
Inverse of a quadratic function with domain restriction

Question 202334: Find the inverse of quadratic function, graph function and its inverse in the same coordinate plane.
y=x^2-2x+1
How do I find the answer? Found 2 solutions by stanbon, Earlsdon:Answer by stanbon(75887) (Show Source):

You can put this solution on YOUR website! 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
—————-
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:
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Cheers,
Stan H.

You can put this solution on YOUR website! To find the inverse of the function: Exchange the x- and y-variables, then solve the resulting equation for y, thus: Solve this equation for y. Subtract X from both sides. Use the quadratic formula to solve. In the equation just developed, a = 1, b = -2, and c = (1-x). Simplify. or which can be further simplified into: This is the inverse.
In order to graph this, you must graph the two solutions as separate graphs: and
The graph looks like:
The red parabola is the graph of the given quadratic equation while the blue & green graphs combine to form the graph of the inverse funtion.

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