1. ## Problem finding highest point in quadratic curve

I have most of this maths puzzle solved but it is the last part which I can't quite work out. Any help would be much appreciated.

Here is the question:

The height h metres of a ball at time t seconds after it is thrown up in the air is given by the expression: h = 1 + 15t - 5t^2
(i) find the times at which the height is 11m.
(ii) Use your calculator to find the time at which the ball hits the ground.
(iii) What is the greatest height the ball reaches?

This is what I have so far:
(i) -5t^2 + 15t + 1 = 11 (height of 11 m).
-5t^2 + 15t -10 = 0
(-5t + 5)(t - 2)
t = +1 or +2
ie in 1 second on the way up and 2 seconds on the way down.

(ii) -5t^2 +15t + 1 = 0 (0 height - ie on ground)
trickier to factorise so use quadratic equation (-b+/-sqrt(b^2-4as))/2a
= -15 +/- 7Sqrt(5)
----------------
-10

t = -0.0652 or 3.0652

Clearly the negative value is not relevant here so we say the ball hits the ground after 3.0652 seconds.

(iii) This is the part of the question which I am struggling on. I know that I need to use complete the square technique. Take coefficient - 15, then divide by two then square it - (15/2)^2 = 225/4 or 56.25

-5t^2 +15t + 1

Add 224/5 to right side, subtract from left side

-5t^2 + 15t + 224/4 1-225/4 = -221/4

I thought the value in the right side - my -221/4, would be the answer - but it is not. What am I doing wrong?

2. To complete the square, the $a$ value needs to be $1$.

$-5t^2 + 15t + 1 = -5\left(t^2 - 3t - \frac{1}{5}\right)$

$= -5\left[t^2 - 3t + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - \frac{1}{5}\right]$

$= -5\left[\left(t - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{1}{5}\right]$

$= -5\left[\left(t - \frac{3}{2}\right)^2 - \frac{45}{20} - \frac{4}{20}\right]$

$= -5\left[\left(t - \frac{3}{2}\right)^2 - \frac{49}{20}\right]$

$= -5\left(t - \frac{3}{2}\right)^2 + \frac{49}{4}$.

3. If $f(x) = ax^2+bx+c$, then the maximum/minimum is at $\left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right)$, where $\Delta = b^2-4ac$.

So for $f(x) = -5t^2 + 15t + 1$, we have $\left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right) = \left (\frac {15}{10}, \frac {15^2+20}{20} \right) = \left(\frac{3}{2}, \frac {245}{20} \right) = \left (\frac {3}{2}, \frac {49}{4} \right).$

4. Originally Posted by TheCoffeeMachine
If $f(x) = ax^2+bx+c$, then the maximum/minimum is at $\left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right)$, where $\Delta = b^2-4ac$.

So for $f(x) = -5t^2 + 15t + 1$, we have $\left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right) = \left (-\frac {15}{10}, \frac {15^2+20}{20} \right) = \frac {245}{20} \right) = \left (-\frac {3}{2}, \frac {49}{4} \right).$
I believe you'll find it's actually $(x,y) = \left(\frac{3}{2}, \frac{49}{4}\right)$...

5. Originally Posted by Prove It
I believe you'll find it's actually $(x,y) = \left(\frac{3}{2}, \frac{49}{4}\right)$...
His working was correct, he just didn't get the final answer correct.

It should have been:

$\left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right) = \left (-\frac {15}{-10}, \frac {15^2+20}{20} \right) = \frac {245}{20} \right) = \left (\frac {3}{2}, \frac {49}{4} \right).$

6. Why does a have to be 1?