What is the inverse of h(t)=3(t-2)^2-4 ?

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

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inverse of h(t)=3(t-2)^2-4

inverse

h (t) = 3 (t - 2)^{2} - 4

\frac{t + 4}{3} + 2, - \frac{t + 4}{3} + 2

Solution steps

y = 3 (t - 2)^{2} - 4

Replace

t

with

y

t = 3 (y - 2)^{2} - 4

Solve for

y, t = 3 (y - 2)^{2} - 4

\frac{t + 4}{3} + 2, - \frac{t + 4}{3} + 2

f (x) = \frac{x - 3}{x + 6}

f (x) = \frac{5}{x + 5}

cos (2 t)

6 x + 9

\frac{1}{4 x - 1}

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