# Solving Word Problems Involving Quadratic Functions

**Solving Word
Problems Involving Quadratic Function**

Some of the most important functions in applications are the quadratic functions. The quadratic functions are one of the simplest type of polynomial function, aside from the linear functions. Applications include satellite dishes, projectile paths, economics, geometry, ecology, weather,biology and many others.

A function f is a quadratic function if f(x)
= ax^{2} + bx +
c where a, b, and c are real numbers and
not equal to 0.. The graph of a quadratic function is a parabola. The most
basic aid in graphing a parabola is knowing
whether a > 0 (the graph opens upward) or a < 0 (the graph opens downward). The
two simplest quadratic functions are f(x) = x^{2
}and g(x) = - x^{2 }.

In this hub, I feature two interesting problems with their complete solution.

Problem Number One :

Jack Pott dives off the high diving board. His distance from the surface of the water

varies quadratically with the number of seconds that have passed since he left the

board.

(a) His distance at time 1, 2 and 3 seconds since he left the board are 24, 18, 2

meters above the water respectively. Write a particular quadratic equation

expressing the distance in terms of time.

(b) How high is the diving board ?

(c ) When does he hit the water ?

Solution :

(a) In finding the quadratic equation three points can be derived from the given :

(1, 24 ) ; (2, 18) ; (3, 2);

by letting the time in seconds as x-coordinates and the distance above the

water as y-coordinates.

The quadratic equation has a standard
form ax^{2} + bx + c
= 0.

Using the given points above, we must solve for a, b, and c.

From the point (1, 24) we can derive an equation by substituting 1 to x’s and

24 to y, The equation formed is : a + b + c = 24 let this be equation #1.

From point (2, 18) we can derive this equation : 4a + 2b + c = 18 let this be

equation #2.

From point ( 3, 2 ) we can derive this equation : 9a + 3b + c = 2 let this be

Equation #3.

We will now solve for a, b, and c using elimination method. by subtraction.

Using equation 1 and equation 2 to eliminate c ;

a + b + c = 24

-4a - 2b - c = -18

__________________________________

-3a - b = 6 let this be equation 4

Using equation 2 and equation 3 to eliminate c

4a + 2b + c = 18

-9a - 3b + c = - 2

-5a - b = 16 let this be equation 5

Solving for a and b using equation 4 and equation 5

-3a –b = 6

+5a +b = -16

-------------------------------------

2a = -10

a = -5

substituting a = -5 in equation 4

-3(-5) – b = 6

15 – b = 6

b = 15 – 6

b = 9

substituting a = -5 b = 9 in equation 1 to solve for c

-5 + 9 + c = 24

c = 24 – 4

c = 20

Therefore the quadratic equation we
are looking for is -5x^{2} + 9x
+ 20 = 0.

(b) How high is the diving board ?

In order to solve for the height
of the diving board we must solve for the vertex of the parabola y = -5x^{2 } + 9x +
20.. This parabola opens downward since a is negative.

Actually, the vertex is the maximum point of the parabola. The height of the diving board is the y-coordinate of the vertex. Solve first for h or x-coordinate of the vertex . The formula to solve for h is -b/2a. Therefore h = -9/(2) (-5) = 9/10.

Substitute 9/10 to every x in
the given quadratic function y = -5x^{2} + 9x + 20.

y = -5 (9/10)^{2 }+ 9 (9/10) + 20

y = -81/20 + 81/10 + 20

y = 481/20 = 24.05 meters.

( c ) When does he hit the water ?

y = 0, when the diver hits the water so in order to solve for this we must let

-5x^{2} +
9x + 20 = 0
and solve for x using quadratic formula.

The quadratic formula is given by :

X = ( -b + - SQRT(b^2-4ac) )/2a

X = (-9 + - SQRT(81 – (4)(-5)(20)) ) /2(-5)

X = (-9 +- SQRT(481) )/-10

X = ( -9 – 21.93 ) /-10

X = -30.3/-10

X = 3.03 seconds

Problem Number Two :

A satellite dish in storage has parabolic cross sections and is resting on its

vertex.. A point on the rim is 4 feet high and is 6 feet horizontally from the

vertex. How high is a point which is 3 feet horizontally from the vertex?

Solution :

Plotting this parabola on a rectangular coordinate plane with vertex at (0, 0 ), we can derive other two points based from the given . Two points on the rim are

(-6, 4) and (6, 4).

Now let us derive the equation of the parabola using points (0, 0) ; (-6, 4), (6, 4).

C = 0

Using point (-6, 4 ) and (6, 4)

4 = a (-6)^2 + (-6)b + c

4 = 36 a -6b + c

4 = 36 a + 6b + c

========================

8 = 72 a

a = 1/9

Solving for b ;

4 = 36 (1/9) + 6b

4 = 4 + 6b

B = 0

The equation of the parabola is y = 1/9 x^2

Y = 1/9 (3)^2

Y = 1/9(9) = 1

The point which is 3 feet horizontally from the vertex is 1 foot high.

## Comments

Cristina, I have to admit math was never one of my better subjects. Glad you have it all together.

Wow! I'm glad there are brains like yours that are big enough to wrap around quadratic word problems - thanks for sharing here!

Thank you Cristina for carrying me out of my comfort zone into a realm of which I know very little. Quite interesting if not challenging.

God bless you,

Forever His,

Ate Cristina, thanks for this Hub. I'm afraid you lost me though for when it comes to math 1 + 1 = 11 I get lost. I do appreciate learning though.

KUYA Dave.

thanks!

this hub is si nice.:)

it helps me a lot .

thanks for creating this beautiful hub :)

--

more blessings 2 come . god bless.

very good answers

I'm glad because only few Filipinos are blogging about math. I also have mine at http://mathandmultimedia.com/.

Keep up the good work Cristina.

Cristina, great great and once again great! That is what I'm looking for! :) Thanks!

I love math

Given Problem:

A cellular phone store can sell 100 phonekits per month for P5,000 each. Reseach indicates that the store can sell 50 more kits per month for each P100 decrease in price. What price per kits per month give greatest monthly sales? What is the sales amount?

Can you explain the answer and what is the mean of "50 more and P100 decrease".

for me nice subject that:P)

When i employed to find on top of existence although recently We've established a new opposition.

I don't think your height of the diving board is correct. You found the max height the diver gets above the water. Wouldn't the diving board height be when t = 0? I think the diving board is 20 meters and the diver reaches the height of 24.05 meters/.9 seconds after leaving the board.

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