What is the inverse of y=3(x+2)^2-19 ?

The inverse of y=3(x+2)^2-19 is sqrt((x+19)/3)-2,-sqrt((x+19)/3)-2

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inverse of y=3(x+2)^2-19

inverse

y = 3 (x + 2)^{2} - 19

\frac{x + 19}{3} - 2, - \frac{x + 19}{3} - 2

Solution steps

y = 3 (x + 2)^{2} - 19

Replace

x

with

y

x = 3 (y + 2)^{2} - 19

Solve for

y, x = 3 (y + 2)^{2} - 19

\frac{x + 19}{3} - 2, - \frac{x + 19}{3} - 2

What is the inverse of y=3(x+2)^2-19 ?
The inverse of y=3(x+2)^2-19 is sqrt((x+19)/3)-2,-sqrt((x+19)/3)-2