1. Quadratics in Real Life

2. Quadratic functions are more than algebraic curiosities—they are widely used in science, business, and engineering. The U-shape of a parabola can describe the trajectories of water jets in a fountain and a bouncing ball, or be incorporated into structures like the parabolic reflectors that form the base of satellite dishes and car headlights. Quadratic functions help forecast business profit and loss, plot the course of moving objects, and assist in determining minimum and maximum values. Most of the objects we use every day, from cars to clocks, would not exist if someone, somewhere hadn't applied quadratic functions to their design.

3. We commonly use quadratic equations in situations where two things are multiplied together and they both depend of the same variable.
For example, when working with area, if both dimensions are written in terms of the same variable, we use a quadratic equation.
Because the quantity of a product sold often depends on the price, we sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold.
Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.

4. Balls, Arrows, Missiles and Stones
If you throw a ball (or shoot an arrow, fire a missile or throw a stone) it will go up into the air, slowing down as it goes, then come down again ...
... and a Quadratic Equation tells you where it will be!

5. Example: Throwing a Ball
A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?
Ignoring air resistance, we can work out its height by adding up these three things:
The height starts at 3 m:   3
It travels upwards at 14 meters per second (14 m/s):   14t
Gravity pulls it down, changing its speed by about 5 m/s per second (5 m/s2):   -5t2  

6. Add them up and the height h at any time t is:
h=-5t2 + 14t + 3
And the ball will hit the ground when the height is zero:
-5t2 + 14t + 3 = 0
It will be easier to solve if we factor out the negative 1:
-1(5t t - 3 ) = 0
Divide both sides by -1 and we have…

7. 5t t - 3 = 0
5t2 - 14t – 3 = (5t + 1)(t – 3) So our two solutions are: t = -0.2 and t = 3 The "t = -0.2" is a negative time, impossible in our case. The "t = 3" is the answer we want: The ball hits the ground after 3 seconds!

8. What does it look like? It shows you the height of the ball vs time
Here is the graph of the Parabola h = -5t2 + 14t + 3
It shows you the height of the ball vs time
Some interesting points:
(0,3) When t=0 (at the start) the ball is at 3 m
(-0.2,0) Says that -0.2 seconds BEFORE we threw the ball it was at ground level ... this never happened, so our common sense says to ignore it!
(3,0) Says that at 3 seconds the ball is at ground level.
Note also that the ball reaches nearly 13 meters high.

9. What about that maximum?
You can find exactly where the top point is!
Find where (along the horizontal axis) the top occurs using -b/2a:
t = -b/2a = -(-14)/(2 × 5) = 14/10 = 1.4 seconds
Then find the height using that value (1.4)
h = -5t2 + 14t + 3 = -5(1.4) × = 12.8 meters
So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

10. Let’s take a look at some more examples…
Profit