What is the inverse of y=4x^2-8x+1 ?

The inverse of y=4x^2-8x+1 is (2+sqrt(x+3))/2 ,(2-sqrt(x+3))/2

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inverse of y=4x^2-8x+1

inverse

y = 4 x^{2} - 8 x + 1

\frac{2 + x + 3}{2}, \frac{2 - x + 3}{2}

Solution steps

y = 4 x^{2} - 8 x + 1

Replace

x

with

y

x = 4 y^{2} - 8 y + 1

Solve for

y, x = 4 y^{2} - 8 y + 1

\frac{2 + x + 3}{2}, \frac{2 - x + 3}{2}

y = 3 lo g_{10} (x - 1) + 4

f (x) = \frac{9 - 5 cos ( 3 x )}{2}

f (x) = 13 - 3 x^{3}

f (x) = \frac{1}{6} x + 12

f (x) = \frac{( 2 x - 1 )}{x ( x - 1 )}

What is the inverse of y=4x^2-8x+1 ?
The inverse of y=4x^2-8x+1 is (2+sqrt(x+3))/2 ,(2-sqrt(x+3))/2