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

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

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

inverse

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

\frac{x + 1}{5} - 4, - \frac{x + 1}{5} - 4

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x + 1}{5} - 4, - \frac{x + 1}{5} - 4

s l o p e y = 7 - 4 x

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

d i s t an ce (\frac{1}{4}, 5), (7, \frac{2}{3})

a sy m pt o t es f (x) = \frac{2}{x - 1} + 3

in t erce pt s cos (2 x + 5)

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