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.

2. Originally Posted by abc10
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 $\displaystyle t = \frac{3}{5}$.

3. To find the y-intercept of a function f(x), simply find f(0). So in your case, what is h(0).

To find extremum without calculus? Hmmm. I'm sure there is a way other than guess and check or running a program. But I'm not remembering.

4. ## Solutions

maximum height of 1.58 y-intercept of 3

5. Originally Posted by nightrider456
maximum height of 1.58 y-intercept of 3
Wrong. The maximum height is actually $\displaystyle \frac{24}{5} = 4.8$.