What is the inverse of y=x^2-4x,x<= 0 ?

The inverse of y=x^2-4x,x<= 0 is 2+sqrt(x+4),2-sqrt(x+4)

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inverse of y=x^2-4x,x<= 0

inverse

y = x^{2} - 4 x, x \leq 0

2 + x + 4, 2 - x + 4

Solution steps

y = x^{2} - 4 x

Replace

x

with

y

x = y^{2} - 4 y

Solve for

y, x = y^{2} - 4 y

2 + x + 4, 2 - x + 4

f (x) = 2 \cdot 1 0^{x}

Q (x) = 45 - 0.45 x

f (x) = \frac{- x - 14}{11}

h (x) = x

What is the inverse of y=x^2-4x,x<= 0 ?
The inverse of y=x^2-4x,x<= 0 is 2+sqrt(x+4),2-sqrt(x+4)