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## PowerPoint Slideshow about 'Quadratic Word Problems' - alaire

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h(t) = -5t2 + 20t + 1, where h is the height of the ball in metres and t is the time in seconds since the ball was released.

h(t) = -5t2 + 20t + 1, where h is the height of the ball in metres and t is the time in seconds since the ball was released.

Probe

Below are 6 different representations of 3 different quadratic relationships. Pair them up, and name the 6 forms.

Probe

Below are 6 different representations of 3 different quadratic relationships. Pair them up, and name the 6 forms.

equation in general form

equation in transformational form

mapping rule

graphical (parabola)

equation in factored form

table of values

Recap

You should know the following about quadratic functions:

- How to graph them
- How to find the vertex
- How to find the x- and y-intercepts
- How to find the equation from the pattern
- How to find the equation from the graph
- How to change from one form to another

Quadratic Word Problems

There are basically two types of quadratic word problems:

Those that ask you to find the vertex

By giving you the equation

Those that ask you to find the roots

Those that ask you to find the vertex

By giving you the information to find the equation

Those that ask you to find the roots

(…these are the harder ones)

Max/Min Values

We have seen that the vertex of a quadratic in general form (y = ax2 + bx + c) is given by:

The y-value of the vertex is either the maximum value the function can have or it’s the minimum value the function can have.

y-value has a min when a > 0

y-value has a max when a < 0

A small business’ profits over the last year have been related to the price of the its only product. The relationship isR(p) = −0.4p2 + 64p − 2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars.

a. What price would maximize the revenue?

The word maximize screams:

“FIND THE VERTEX!!”

Thus, a price of $80 would maximize the revenue.

A small business’ profits over the last year have been related to the price of the its only product. The relationship isR(p) = −0.4p2 + 64p − 2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars.

b. What is the maximum revenue possible?

The answer to this question is the y-value (R-value) of the vertex.

The maximum revenue is $160 000.

A small business’ profits over the last year have been related to the price of the its only product. The relationship isR(p) = −0.4p2+ 64p −2400, where R is the revenue measured in thousands of dollars and p is the price of the product measured in dollars.

c. How much money would they lose if they gave the product away?

This question is talking about a price of 0 or p = 0

The business would lose $2 400 000.

Max/Min Word Problems

Example 2: (no equation given)

A farmer needs to fence off his animals. He bought 800 m of fencing and would like to maximize the total enclosed area. According to regulations the pens need to find these relative dimensions. What is the maximum area he can enclose, and what are its dimensions?

Again the word MAXIMUM means VERTEX…but…

we don’t have an equation!

Hint: Start with the value you are trying to maximize/minimize

kids pen

pig pen

sheep

pen

Max/Min Word Problems

Example 2: (no equation given)

A farmer needs to fence off his animals. He bought 800 m of fencing and would like to maximize the total enclosed area. According to regulations the pens need to find these relative dimensions. What is the maximum area he can enclose, and what are its dimensions?

Using the hint we get:

Area of a rectangle:

A = (3w)(8L)

A = 24wL

kids pen

pig pen

sheep

pen

But there are too many variables!(A, wand L).

Hint: Find an equation using both L and w, and use substitution to eliminate one of them

The total fencing is 800m. This means

800 = 3w+3w+3w+8L+8L+w+2L

800 = 10w + 18L

orw = 80 – 1.8L

Sub it in!

Max/Min Word Problems

Example 2: (no equation given)

A farmer needs to fence off his animals. He bought 800 m of fencing and would like to maximize the total enclosed area. According to regulations the pens need to find these relative dimensions. What is the maximum area he can enclose, and what are its dimensions?

Area of a rectangle:

A = 24wL

A = 24(80 – 1.8L)L

A = 1920L –43.2L2

kids pen

pig pen

sheep

pen

Now we know that the maximum area occurs whenL= 22.22m.

This looks very promising because there are only two variables (A and L) and it is a quadratic (there is an L2)

Now find the vertex! (Because it is asking for MAXimum area)

Max/Min Word ProblemsA farmer needs to fence off his animals. He bought 800 m of fencing and would like to maximize the total enclosed area. According to regulations the pens need to find these relative dimensions. What is the maximum area he can enclose, and what are its dimensions?

Example 2: (no equation given)

A= 1920L – 43.2L2

kids pen

w = 80 – 1.8L

pig pen

sheep

pen

Now we can also find the dimensions:

Now we can find this maximum area:

L = 22.22m

w= 80 – 1.8L

= 80 – 1.8(22.22)

= 40m

Maximum area:

A = 1920(22.22)- 43.2(22.22)2A = 21 333m2

Your Turn

A lifeguard has 75m of rope to section off the supervised area of the beach. What is the largest rectangular swimming area possible?

2w + L = 75

A = wL

L = 75 – 2w

A = w(75 – 2w)

w

A = 75w – 2w2

L

w

Maximum area:

A = 75(18.75) – 2(18.75)2

A =1406.25 – 703.125

A = 703.125m2

Roots of a Quadratic

We have 3 ways to find the roots of (solve) a quadratic equation:

1. Factoring

(sometimes fast, sometimes not possible)

2. Completing the square

(always possible, takes too long)

3. Using the quadratic root formula

(always possible, pretty quick)

Roots Word Problems

Example 3(equation given)

A duck dives under water and its path is described by the quadratic function y= 2x2– 4x, where y represents the position of the duck in metres and x represents the time in seconds.

a. How long was the duck underwater?

The duck is no longer underwater when the depth is 0. We can plug in y = 0 and solve for x.

The duck was underwater for 2 – 0, or 2 seconds

So x = 0 or 2

Example 3(equation given)

A duck dives under water and its path is described by the quadratic function y= 2x2– 4x, where y represents the position of the duck in metres and x represents the time in seconds.

b. When was the duck at a depth of 5m?

We cannot solve this because there’s a negative number under the square root.

We conclude that the duck is never 5m below the water.

Let’s check by finding the minimum value of y.

We can plug in y = -5 and solve for x.

Example 3(equation given)

A duck dives under water and its path is described by the quadratic function y= 2x2– 4x, where y represents the position of the duck in metres and x represents the time in seconds.

c. How long was the duck at least 0.5m below the water’s surface?

We can plug iny= -0.5 and solve for x. This will give us the times when the duck is at exactly 0.5m below.

The duck was exactly 0.5m below at x= 0.14s and at x= 1.87s. Therefore it was below 0.5m for 1.73s

Roots Word Problems

Example 4: (no equation given)

A rectangular lawn measures 8m by 6m. The homeowner mows a strip of uniform width around the lawn, as shown. If 40% of the lawn remains un-mowed, what is the width of the strip?

So 19.2 = (8 – 2x)(6 – 2x)

-X-

-X-

-6 – 2x-

40%

-----8 – 2x-----

The total area of the lawn is 8 ×6 = 48

This is a quadratic equation! What a perfect time to use the Quadratic Root Formula to solve for x.

40% of this is unmowed: 48 ×0.40 = 19.2

The dimensions of this un-mowed rectangle are: 8 – 2x and 6 – 2x.

Roots Word Problems

Example 4: (no equation given)

A rectangular lawn measures 8m by 6m. The homeowner mows a strip of uniform width around the lawn, as shown. If 40% of the lawn remains un-mowed, what is the width of the strip?

-6 – 2x-

-----8 – 2x-----

The mowed strip has a width of 1.25m

Roots Word Problems

Example 4: (no equation given)

A rectangular lawn measures 8m by 6m. The homeowner mows a strip of uniform width around the lawn, as shown. If 40% of the lawn remains un-mowed, what is the width of the strip?

Let’s check:

If x = 1.25m then the

-6 – 2x-

length is

8 – 2x

= 8 – 2(1.25)

= 5.5m

width is

6 – 2x

= 6 – 2(1.25)

= 3.5m

-----8 – 2x-----

A = (5.5)(3.5)

A =19.2

Which was 40% of the total area!

Example 5: (no equation given)

Two numbers have a difference of 18. The sum of their squares is 194.What are the numbers?

Let s represent one of the numbers

Let g represent the other

Since there is no diagram, we have to define our variables

The question indicates two equations relating these two variables.

Again, we have a substitution situation (like example 2). Solve the simpler equation for a variable and plug it in to the other equation.

We can turn this into a quadratic equation in general form. Let’s do it because then we can use the Quadratic Root Formula to solve for g.Let’s go!

s = 18 +g

s – g = 18

s2 + g2 = 194

(18 + g)2 + g2 = 194

324 + 36g + g2 + g2 = 194

2g2 + 36g + 324 - 194 =0

2g2 + 36g + 130 = 0

Example 5: (no equation given)

Two numbers have a difference of 18. The sum of their squares is 194.What are the numbers?

Now apply the Quadratic Root Formula to solve for g.

Now to find s.

2g2 + 36g + 130 = 0

s = 18 + (-5) = 13

ors = 18 + (-13) = 5

Therefore the two numbers are:-5 and 13,or5 and -13

----6---

A rectangle is 8 feet long and 6 feet wide. If each dimension increases by the same number of feet, the area of the new rectangle formed is 32 square feet larger than the area of the original rectangle.How many feet increased each dimension?

--x--

By the Quadratic Root Formula:

By factoring and the zero property:

0 = (x – 2)(x + 16)

x= 2 or x = -16

--x--

The new area is:32 + (6 × 8) = 32 + 48 = 80

The dimensions of the new rectangle are6 + x and 8 + x

80 = (6 + x)(8 + x)

80 = 48 + 14x + x2

0 = x2 + 14x – 32

Therefore, each dimension increased by 2 feet.

Mixin’ it up

A ball is thrown and follows the path described by the function

h(t) = -5t2 + 20t + 1, where h is the height of the ball in metres and t is the time in seconds since the ball was released.

a) When was the ball at a height of 3.5m?

3.5 = -5t2+ 20t + 1

This question is looking for t so it gives a specific h. In this case h = 3.5.

We will solve this by simplifying and using the quadratic root formula

One time is on the way up and the other is on the way down.

0 = -5t2 + 20t– 2.5

The ball is at a height of 3.5m twice: once at t = 0.129sand again at t = 3.871s

A ball is thrown and follows the path described by the function

h(t) = -5t2 + 20t + 1, where h is the height of the ball in metres and t is the time in seconds since the ball was released.

b) How high is the ball after 4.0s?

h = -5(4)2 + 20(4) + 1

h = -80 + 80 + 1

h = 1

This question is looking for h given the value of t = 4.0

After 4.0 seconds in the air, the ball is 1m off the ground.

A ball is thrown and follows the path described by the function

h(t) = -5t2 + 20t + 1, where h is the height of the ball in metres and t is the time in seconds since the ball was released.

c) What is the ball’s maximum height?

The question is asking for height so I must know the time. Do I?

The word MAXIMUM screams VERTEX!! So I do know the time value…

The ball reaches its maximum height 2.0 seconds after being thrown

What is this height?

Therefore, the ball’s maximum height is 21m, which occurs at 2s.

A ball is thrown and follows the path described by the function

d) When does the ball hit the ground?

0 = -5t2 + 20t + 1

This question is asking for the time so I must know the height… h = 0 of course!

Time can’t be negative so this cannot be an answer.

Therefore, the ball h it’s the ground in 4.049 seconds.

A ball is thrown and follows the path described by the function

e) From what height was the ball thrown?

This question is asking for the height so I must know the time…t = 0 of course!

h = -5(0)2 + 20(0)+ 1

h = 1

Therefore, the ball was thrown from a height of 1m.

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