What is the inverse of h(t)=-16t^2+32t+5 ?

The inverse of h(t)=-16t^2+32t+5 is -(-4+sqrt(-t+21))/4 ,(sqrt(-t+21)+4)/4

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inverse of h(t)=-16t^2+32t+5

inverse

h (t) = - 16 t^{2} + 32 t + 5

- \frac{- 4 + - t + 21}{4}, \frac{- t + 21 + 4}{4}

Solution steps

y = - 16 t^{2} + 32 t + 5

Replace

t

with

y

t = - 16 y^{2} + 32 y + 5

Solve for

y, t = - 16 y^{2} + 32 y + 5

- \frac{- 4 + - t + 21}{4}, \frac{- t + 21 + 4}{4}

x^{2} + 8 x - 3

2.5 t + 5.5

f (x) = \frac{( x + 4 )}{( x + 10 )}

f (x) = - x - 1 - 1

f (x) = arcsin (2 x - 1)

What is the inverse of h(t)=-16t^2+32t+5 ?
The inverse of h(t)=-16t^2+32t+5 is -(-4+sqrt(-t+21))/4 ,(sqrt(-t+21)+4)/4