What is the inverse of h(x)=-3(x+4)^2-1 ?

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

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inverse of h(x)=-3(x+4)^2-1

inverse

h (x) = - 3 (x + 4)^{2} - 1

- \frac{x + 1}{3} - 4, - - \frac{x + 1}{3} - 4

Solution steps

y = - 3 (x + 4)^{2} - 1

Replace

x

with

y

x = - 3 (y + 4)^{2} - 1

Solve for

y, x = - 3 (y + 4)^{2} - 1

- \frac{x + 1}{3} - 4, - - \frac{x + 1}{3} - 4

\frac{- 3 + 4 x - 3}{2}

\frac{6 x - 6}{6 x + 1}

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

f (x) = (32)^{5}

f (x) = x^{4} + 4, x \leq 0

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