What is the inverse of h(t)=-16t^2+48t+5 ?

The inverse of h(t)=-16t^2+48t+5 is -(-6+sqrt(-t+41))/4 ,(sqrt(-t+41)+6)/4

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inverse of h(t)=-16t^2+48t+5

inverse

h (t) = - 16 t^{2} + 48 t + 5

- \frac{- 6 + - t + 41}{4}, \frac{- t + 41 + 6}{4}

Solution steps

y = - 16 t^{2} + 48 t + 5

Replace

t

with

y

t = - 16 y^{2} + 48 y + 5

Solve for

y, t = - 16 y^{2} + 48 y + 5

- \frac{- 6 + - t + 41}{4}, \frac{- t + 41 + 6}{4}

(\frac{3 - 2 x}{2 x - 5}) - 1

f (x) = \frac{x + 13}{5}

\frac{3 x + 2}{x ^{2} - 7 x}

y = lo g_{10} (3 x)

f (x) = (x + 4) 2 - 3

What is the inverse of h(t)=-16t^2+48t+5 ?
The inverse of h(t)=-16t^2+48t+5 is -(-6+sqrt(-t+41))/4 ,(sqrt(-t+41)+6)/4