What is the inverse of f(x)=-1/4 (x+3)^2-5 ?

The inverse of f(x)=-1/4 (x+3)^2-5 is 2sqrt(-x-5)-3,-2sqrt(-x-5)-3

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inverse of f(x)=-1/4 (x+3)^2-5

inverse

f (x) = - \frac{1}{4} (x + 3)^{2} - 5

2 - x - 5 - 3, - 2 - x - 5 - 3

Solution steps

y = - \frac{1}{4} (x + 3)^{2} - 5

Replace

x

with

y

x = - \frac{1}{4} (y + 3)^{2} - 5

Solve for

y, x = - \frac{1}{4} (y + 3)^{2} - 5

2 - x - 5 - 3, - 2 - x - 5 - 3

x^{4} - 6 x^{3}

(0, 0), (\frac{1}{2}, 1)

f (x) = \frac{x + 2}{x ^{2} - 17 x + 72}

f (x) = \frac{3 x + 1}{x ^{2} + x - 2}

\frac{1}{cos ( x )}

What is the inverse of f(x)=-1/4 (x+3)^2-5 ?
The inverse of f(x)=-1/4 (x+3)^2-5 is 2sqrt(-x-5)-3,-2sqrt(-x-5)-3