Affine functions : corrected math exercises in 3ème.

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Answers to math exercises in 3rd grade on affine functions in PDF. Know how to draw the curve of an affine function and determine, by calculation, the value of the directing coefficient bet the y-intercept in third.

Exercise 1:

Explain what the following notations mean:

a. f : x 3x+7 : the function f that associates the number 3x+7 to any number x.
b. f(x)= -2x+3 : the function f defined by the image of any number x is -2x+3

Exercise 2:
Among the given functions, indicate which are affine, which are linear, and which are not affine.

is affine .

is affine.

is linear

is affine, it is a constant function.

is neither linear nor affine.

$k(x)=-\frac{3x}{7}=-\frac{3}{7}x$ is linear.

$l(x)=3\sqrt{x}$ is neither linear nor affine.

$m(x)=3+\frac{1}{x}$ is neither linear nor affine.

Exercise 3:

The function f is defined by : x -5x+2.

a. Compute f(2)=-8 ;f(- 3)=17 ; f(0)=2.
b. Calculate the image of 4: f(4)= -18
c. Calculate the number x such that :
$f(x)=\frac{5}{3}$ .

$-5x+2=\frac{5}{3}$

$-5x=-2+\frac{5}{3}$

$-5x=-\frac{6}{3}+\frac{5}{3}$

$-5x=-\frac{1}{3}$

$x=\frac{1}{15}$

Exercise 4:

We give the images of two numbers by an affine function f.

f(3)=5 and f(7)=13

a. Draw its graphical representation in a reference frame.

b. Determine the algebraic expression of this function f : x ax+b (i.e. determine a and b).

$a=\frac{f(7)-f(3)}{7-3}=\frac{13-5}{4}=\frac{8}{4}=2$

$f(3)=5\Leftrightarrow,f(3)=2\times ,3+b=5$

Conclusion: ${\color{DarkRed},f(x)=2x-1}$

Exercise 6:

g(x)=3x

u(x)=2

f(x)=x+2.5

h(x)=-2x-2

k(x)=-3.5x+0.5

Exercise 7:

The lines are parallel because the directrix is the same.

Exercise 8:

a. For rate 1, he will pay 100 €.

For rate 2, he will pay: 40+12×1=52 €.

For rate 3, he will pay: 12×2= 24 €.

In this case, rate 3 is the most interesting.

b. We call x the number of times that Yéro will go to the pool.

Express, as a function of x :

_t1(x)=100

_t2(x)= x+40

_t3(x)= 2x

c. Graphically represent these three functions in the same orthogonal reference frame.

d. He will have to pay 12×4 = 48 entries.

What if he goes twice a week?

double that is 96 entries.

e. For 48 entries, it is rate 2 and for 96 it is rate 1.

f. From 61 entries on, rate 1 is the most interesting.

Exercise 9:

1)Be the affine function f defined by f(x)=-2x+3 .

a) calculate f(0) .

f(0)= -2×0+3=3

b) calculate the antecedent of 5 .

fx()=5

-2x+3=5

-2x=5-3

-2x=2

x=-1

Therefore the antecedent of 5 by the function f is – 1 .

2) Let the affine function g be such that g(-2)=-2 and g(3)=4 .

a) determine the function g .

$a=\frac{g(3)-g(-2)}{3-(-2)}=\frac{4-(-2)}{5}=\frac{6}{5}$

and

$g(-2)=-2\times \,\frac{6}{5}+b$

$g(-2)=-\frac{12}{5}+b$

$-2=-\frac{12}{5}+b$

$b=-2+\frac{12}{5}$

$b=-\frac{10}{5}+\frac{12}{5}$

${\color{DarkRed}\,b=-\frac{2}{5}}$

Conclusion : the affine function g is

${\color{DarkRed}\,g(x)=\frac{6}{5}}x-\frac{2}{5}$

b) calculate g(0) and g(3) .

$g(0)=\frac{6}{5}\times \,0-\frac{2}{5}=-\frac{2}{5}$

$g(3)=\frac{6}{5}\times \,3-\frac{2}{5}=\frac{18}{5}-\frac{2}{5}=\frac{16}{5}$

3) in the same reference frame (O,I,J).

a) draw the graphical representations of f and g .

b) calculate the coordinates of the point of intersection of these graphs.

Let’s solve for f(x) = g(x)

$-2x+3=\frac{6x-2}{5}$

$x=\frac{-17}{-16}=\frac{17}{16}$

and

$f(\frac{17}{16})=-2\times \,\frac{17}{16}+3=-\frac{17}{16}+\frac{48}{16}=\frac{31}{16}$

The coordinates of the intersection point are: $I(\frac{17}{16};\frac{31}{16})$ .

Exercise 10:

For the payment of the daycare in a school, two formulas are proposed.

– Formula A : you pay 40 € to become a member for the school year then you

pays 10 € per month for daycare.

Formula B: for non-members, we page 18 € per month.

1. The number of months of daycare is .

and

2. Graph the following functions in a reference frame:

$x,\mapsto ,A(x)=10x+40$ and $x,\mapsto ,B(x)=18x$ .

We will take 1 cm for 1 month in abscissa and 1 cm for 10 € in ordinate.

a) This is the point of intersection of the two lines. For 9 months, the prices to be paid are equal.

b) Find this result by calculation.

$x=\frac{40}{8}$

${\color{DarkRed},x=5}$

4. The most advantageous if you only pay for 4 months in a year is formula B.

5. We have a budget of 113 €.

We have to solve this equation:

$A(x)\le\,113\\10x+40\le\,113\\10x\le\,113-40\\10x\le\,73\\x\le\,\frac{73}{10}\\x\le\,7,3$

We will be able to pay for a maximum of 7 months of daycare with a budget of 113 euros.

Exercise 11:

In each of the following cases, write the function f as f(x)=ax+b

and specify the values of a and b.

1) The graphical representation of f is a straight line of direction -3 and such that f(0)=2.

a= – 3 so f(x)=-3x+b

and f(0)= – 2

therefore -3×0+b= – 2

b= – 2

so f (x)= – 3x – 2

2)The function f is the function that adds 6 to a number x and multiplies the result by – 4.

f(x)= – 4(x+6) = -4x-24

3) The function f is the function that, to a number x, multiplies it by 3, adds 4 to the result,

then divide it by 2.

$f(x)=\frac{3x+4}{2}=\frac{3}{2}x+\frac{4}{2}={\color{DarkRed}\,1,5x+2}$

4) The function f is defined by f(x)=(x+1)²-x².

${\color{DarkRed}\,f(x)=2x+1}$

5). The function f is such that if the x’s increase by 3, the “f(x)” increases by 12.

Moreover, f(0)=1.

$a=\frac{12}{3}=4$

and f(x)=4x+b

moreover f(0)=4×0+b=1

so b=1

where f(x) = 4x+1

Exercise 12:

We designate by the height SK (expressed in meters) of the SABCD pyramid.

1) Show that the volume (in^m3) of the greenhouse is given by the formula $V\,=\,144\,+\,16x$
x
.

$V=8\times \,6\times \,3+\frac{1}{3}\times \,8\times \,6\times \,x$

${\color{DarkRed}\,V=148+16\,x}$

2) Calculate this volume for
x
= 1,5.

$V=148+16\,x=148+16\times \,1,5=148+24=172$

3) For what value of is the volume of the greenhouse 200^m3?

$x=\frac{52}{16}$

The height of the pyramid must be 3.25 m so that the volume is 200 .

Exercise 13:

1. Complete the table.

2. a.

a. For 6 cartridges, price A is the most advantageous.

b. For 80 euros, it is more interesting to choose formula B.

Let’s find out for which number of cartridges the prices are equal:

$x=\frac{40}{5}$

Prices are equal for 8 cartridges purchased.

The internet price is lower for more than 8 cartridges.

Exercise 14:

The school decides to test a software to manage its library. She downloads this software on
Internet.

1. The file is 3.5 MB (Mega-bytes) in size and takes 7 seconds to download.

What is the speed of the Internet connection? The result will be given in MB/s.
it is a situation of proportionality.

$3,5Morightarrow\,7s$
$xrightarrow\,1s$

$x=\frac{3,5\times \,1}{7}=0,5Mo/s$

After a 1 month trial period, the school decides to purchase the software.

There are three rates:

– Rate A: 19 €.
– Rate B: 10 cents per student
– Rate C : 8 € + 5 cents per student

2. Complete the following table:

Number of students	100	200	300
Rate A	19 €
Rate B			30 €
Rate C		18 €

3. a. If x represents the number of students, which of the following expressions corresponds to rate C?
C1 = 8 + 5x
C2 = 8 + 0.05x
C3 = 0.05 + 8x

b. Is this a situation of proportionality? Justify the answer.
C2 corresponds to an affine function, so it is not a situation of proportionality, there should have been a linear function (it is the case of tariff B).
Exercise 17:

a. The total area is :

$10\times \,8-\frac{3\times \,10}{2}=80-15=65\,m^2$ .

b. x is between 0 and 5.

$c(x)=\frac{(5-x+8)\times \,10}{2}=5(13-x)=65-5x$

e. We observe graphically that the wishes of the librarian

will be taken into account when x = 5.

and by calculation :

10x=65-5x

10x+5x=65

15x=65

$x=\frac{65}{15}=5$

The answers to the math exercises on affine functions in 3rd grade.

After having consulted the answers to these exercises on affine functions in 3rd grade, you can return to the exercises in 3rd grade.

Third grade exercises.

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