Here we have collected some examples for you, and solve each using different methods:

Each example follows three general stages:

Take the real world description and make some equations Solve!Use your common sense to interpret the results## Balls, Arrows, Missiles and Stones

When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

You are watching: Real world example of a function

... and a Quadratic Equation tells you its position at all times!

## 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:**(Note: t** is time in seconds)

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 position by about 5 m per second squared: | −5t2 | |

(Note for the enthusiastic: the -5t2 is simplified from -(½)at2 with a=9.8 m/s2) |

Add them up and the height **h** at any time **t** is:

h = 3 + 14t − 5t2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1:

5t2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give **a×c**, and add to give **b**" method in Factoring Quadratics:

a×c = **−15**, and b = **−14**.

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a fewcombinations we find that **−15** and **1** work (−15×1 = −15, and −15+1 = −14)

Rewrite middle with −15 and 1:5t2 − 15t + t − 3 = 0

Factor first two and last two:5t(t − 3) + 1(t − 3) = 0

Common Factor is (t − 3):(5t + 1)(t − 3) = 0

And the two solutions are:5t + 1 = 0 or t − 3 = 0

t =

**−0.2**or 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!

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.

Also notice that the ball goes **nearly 13 meters** high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations, and has two steps:

Find where (along the horizontal axis) the top occurs using **−b/2a**:

**1.4 seconds**

Then find the height using that value (1.4)

h = −5t2 + 14t + 3 = −5(1.4)2 + 14 × 1.4 + 3 =**12.8 meters**

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

## Example: New Sports BikeYou have designed a new style of sports bicycle! Now you want to make lots of them and sell them for profit. |

Your **costs** are going to be:

Based on similar bikes, you can expect **sales** to follow this "Demand Curve":

Where "P" is the price.

For example, if you set the price:

at $0, you just give away 70,000 bikes at $350, you won"t sell any bikes at allat $300 you might sell**70,000 − 200×300 = 10,000**bikes

So ... what is the best price? And how many should you make?

**Let us make some equations!**

How many you sell depends on price, so use "P" for Price as the variable

Unit Sales = 70,000 − 200PSales in Dollars = Units × Price = (70,000 − 200P) × P = 70,000P − 200P2Costs = 700,000 + 110 x (70,000 − 200P) = 700,000 + 7,700,000 − 22,000P = 8,400,000 − 22,000PProfit = Sales-Costs = 70,000P − 200P2 − (8,400,000 − 22,000P) = −200P2 + 92,000P − 8,400,000Profit = −200P2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square.

### Solve: −200P2 + 92,000P − 8,400,000 = 0

**Step 1** Divide all terms by -200

P2 – 460P + 42000 = 0

**Step 2** Move the number term to the right side of the equation:

P2 – 460P = -42000

**Step 3** Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2)2 = (−460/2)2 = (−230)2 = 52900

P2 – 460P + 52900 = −42000 + 52900

(P – 230)2 = 10900

**Step 4** Take the square root on both sides of the equation:

P – 230 = ±√10900 = ±104 (to nearest whole number)

**Step 5** Subtract (-230) from both sides (in other words, add 230):

P = 230 ± 104 = 126 or 334

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don"t we?

**It is exactly half way in-between!** At $230

And here is the graph:

**Profit = −200P2 + 92,000P − 8,400,000**

**The best sale price is $230**, and you can expect:

**$2,180,000**

A very profitable venture.

## Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be **28 cm2**

The inside of the frame has to be** 11 cm by 6 cm**

What should the width **x** of the metal be?

Area of steel before cutting:

Area = (11 + 2x) × (6 + 2x) cm2

Area = 66 + 22x + 12x + 4x2

Area = 4x2 + 34x + 66

Area of steel after cutting out the 11 × 6 middle:

Area = 4x2 + 34x + 66 − 66

### Let us solve this one graphically!

Here is the graph of 4x2 + 34x :

The desired area of **28** is shown as a horizontal line.

The area equals 28 cm2 when:

**x is about −9.3 or 0.8**

The negative value of **x** make no sense, so the answer is:

x = 0.8 cm (approx.)

## Example: River Cruise

### A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat"s speed and how long was the upstream journey?

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

Let**x**= the boat"s speed in the water (km/h)Let

**v**= the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

when going upstream,**v = x−2**(its speed is reduced by 2 km/h)when going downstream,

**v = x+2**(its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by **(x-2)****(x+2)**:

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x2−4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x2 − 30x − 12 = 0

It is a Quadratic Equation!Let us solve it using the Quadratic Formula:

/ 2a">Where **a**, **b** and **c** are from the **Quadratic Equation in "Standard Form": ax2 + bx + c = 0**

### Solve 3x2 - 30x - 12 = 0

**Coefficients are:**

**a = 3**,

**b = −30**and

**c = −12**

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat"s Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

## Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "RT" is:

*1***RT** = *1***R1** + *1***R2**

In this case, we have RT = 2 and R2 = R1 + 3

*1***2** = *1***R1** + *1***R1+3**

To get rid of the fractions we can multiply all terms by 2R1(R1 + 3) and then simplify:

Multiply all terms by 2R1(R1 + 3):

*2R1(R1+3)*

**2**=

*2R1(R1+3)*

**R1**+

*2R1(R1+3)*

**R1+3**

Yes! A Quadratic Equation !

Let us solve it using our Quadratic Equation Solver.

Enter 1, −1 and −6 And you should get the answers −2 and 3 R1 cannot be negative, so **R1**** = 3 Ohms** is the answer.

The two resistors are 3 ohms and 6 ohms.

## Others

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Quadratic equations are also needed when studying lenses and curved mirrors.

See more: Which Instrument Is Most Often Used To Measure Acid Volume Before A Titration Begins?

And many questions involving time, distance and speed need quadratic equations.

Quadratic EquationsFactoring QuadraticsCompleting the SquareGraphing Quadratic EquationsQuadratic Equation SolverAlgebra Index