Find the Function h(t)=-t^2-2t-4

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

The function $H (t)$ can be found by evaluating the indefinite integral of the derivative $h (t)$ .

$H (t) = \int h (t) d t$

Split the single integral into multiple integrals.

$\int - t^{2} d t + \int - 2 t d t + \int - 4 d t$

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$- \int t^{2} d t + \int - 2 t d t + \int - 4 d t$

By the Power Rule, the integral of $t^{2}$ with respect to $t$ is $\frac{1}{3} t^{3}$ .

$- (\frac{1}{3} t^{3} + C) + \int - 2 t d t + \int - 4 d t$

Since $- 2$ is constant with respect to $t$ , move $- 2$ out of the integral.

$- (\frac{1}{3} t^{3} + C) - 2 \int t d t + \int - 4 d t$

By the Power Rule, the integral of $t$ with respect to $t$ is $\frac{1}{2} t^{2}$ .

$- (\frac{1}{3} t^{3} + C) - 2 (\frac{1}{2} t^{2} + C) + \int - 4 d t$

Apply the constant rule.

$- (\frac{1}{3} t^{3} + C) - 2 (\frac{1}{2} t^{2} + C) - 4 t + C$

Simplify.

Combine $\frac{1}{3}$ and $t^{3}$ .

$- (\frac{t^{3}}{3} + C) - 2 (\frac{1}{2} t^{2} + C) - 4 t + C$

Combine $\frac{1}{2}$ and $t^{2}$ .

$- (\frac{t^{3}}{3} + C) - 2 (\frac{t^{2}}{2} + C) - 4 t + C$

Simplify.

$- \frac{t^{3}}{3} - t^{2} - 4 t + C$

Reorder terms.

$- \frac{1}{3} t^{3} - t^{2} - 4 t + C$

The function $H$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$H (t) = - \frac{1}{3} t^{3} - t^{2} - 4 t + C$