What is the inverse of y=(x-3)^2+7,x>3 ?

The inverse of y=(x-3)^2+7,x>3 is sqrt(x-7)+3,-sqrt(x-7)+3

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inverse of y=(x-3)^2+7,x>3

inverse

y = (x - 3)^{2} + 7, x > 3

x - 7 + 3, - x - 7 + 3

Solution steps

y = (x - 3)^{2} + 7

Replace

x

with

y

x = (y - 3)^{2} + 7

Solve for

y, x = (y - 3)^{2} + 7

x - 7 + 3, - x - 7 + 3

f (x) = e^{x}, x \leq 0

\frac{2 x - 3}{5 x + 2}

y = 4 x - 3 + 5

\frac{( z - 8 ) ^{2}}{z ^{2} - 64}

g (x) = \frac{4 x}{7 x - 5}

What is the inverse of y=(x-3)^2+7,x>3 ?
The inverse of y=(x-3)^2+7,x>3 is sqrt(x-7)+3,-sqrt(x-7)+3