Alright, I know this, but its been a long semester, haha.
$\displaystyle h=-5t^2+6t+3 $
What is the maximum height, how long does it take, what is the y intercept.
There won't be a $\displaystyle y$ intercept, as there is no $\displaystyle y$ in your equation.
I think you mean the $\displaystyle h$ intercept.
Let $\displaystyle t = 0$.
Then $\displaystyle h = -5(0)^2 + 6(0) + 3$
$\displaystyle h = 3$.
Since the coefficient of $\displaystyle t^2$ is negative, you WILL have a maximum. To find it, put the equation in Turning Point Form by Completing the Square.
$\displaystyle h = -5t^2 + 6t + 3$
$\displaystyle h = -5\left(t^2 - \frac{6}{5}t - \frac{3}{5}\right)$
$\displaystyle h = -5\left[t^2 - \frac{6}{5}t + \left(-\frac{3}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 - \frac{3}{5}\right]$
$\displaystyle h = -5\left[\left(t - \frac{3}{5}\right)^2 - \frac{24}{25}\right]$
$\displaystyle h = -5\left(t - \frac{3}{5}\right)^2 + \frac{24}{5}$.
So the Turning point is $\displaystyle (t, h) = \left(\frac{3}{5}, \frac{24}{5}\right)$.
So the maximum value of $\displaystyle h$ is $\displaystyle \frac{24}{5}$ which occurs when $\displaystyle t = \frac{3}{5}$.