Domain is $left{-5,-3,-1,1 right}$

Range is $left{ 1,2,3,2right}$

This relation is a function because each element of the domain corresponds with only one element in the range

b) Domain is $left{-1,1,3,5 right}$

Range is $left{ -3,-1,0,2right}$

The value 1 of the independent variable maps
to two different values -3 and 0 of the dependent variable
This relation is not a function.

Domain is $left{0,3,5 right}$

Range is $left{ 4,5,-2,1right}$

The value 0 of the independent variable maps
to two different values 4 and 1 of the dependent variable
This relation is not a function.

Domain is $left{-4,-2,2,5 right}$

Range is $left{ 1,6right}$

Each element of the domain has only one
corresponding element in the range, so the relation
is a function.

a) function ; each $x$-value has a unique $y$-value

b) not a function ; for $x$=1, $y=0$ and $-3$

c) not a function; for $x=0$ , $y=1$ and 4

d) function ; each $x$-value has a unique $y$-value

To determine whether a graph represents a function or not, we can do a $bold{vertical;line; test}$.

If the vertical line touches the graph $bold{only;once}$, then it is a function. If it touches the vertical line $bold{more;than;once}$, it is not a function.

In this case, if you draw a vertical line at $x=1$, it touches the graph at three points. Thus, it is $boxed{not;a;function}$

In this case, if you draw vertical lines at any $x>-4$, the line will touch the graph at two points. Thus, it is $boxed{not;a;function}$.

If you draw any vertical line, it touches the graph only once. Hence, it is a $boxed{function}$.

If you draw a vertical line within the interval $-5leq xleq 1$, it touches the graph at two points. Hence, it is $boxed{not;a;function}$.

Any vertical line would touch the graph only at one point. It is therefore a $boxed{function}$.

If you draw vertical lines where $x<3$,
it would touch the graph at two points. It is therefore $boxed{not;a;function}$

a) List of ordered pairs for the relation between students and grades is

$$
left{ (Barbara, 10), (Pierre, 12), (Kateri,11), (Mandeep, 11), (Elly,10 )right}
$$

b) Use a a mapping diagram for the relation between students and grades we see:

Domain is $left{ Barbara, Pierre, Kateri, Mandeep, Ellyright}$

Range is: $left{10,11,12 right}$

c) Each element of the domain has only one
corresponding element in the range, so the relation student-grade
is a function.

a) List of ordered pairs for the relation between grades and numbers of credits is

$$
left{ (10,8), (12,25), (11,15), (11,18), (10,16 )right}
$$

b) Use a a mapping diagram for the relation between students and grades we see:

Domain is $left{ 10,11,12right}$

Range is: $left{8, 15, 16, 18, 25 right}$

c) Element 11 of the domain has two
corresponding element in the range 15 and 18,and e lement 10 of the domain has two
corresponding element in the range 8 and 16 so the relation grade-credits is not a function.

a) List of ordered pairs for the relation between students and grades is

$$
left{ (Barbara, 8), (Pierre, 25), (Kateri,15), (Mandeep, 18), (Elly,16 )right}
$$

b) Use a a mapping diagram for the relation between students and grades we see:

Domain is $left{ Barbara, Pierre, Kateri, Mandeep, Ellyright}$

Range is: $left{8,15,16,18,25 right}$

c)Each element of the domain has only one
corresponding element in the range, so the relation student-credits is a function.

a) see diagrams inside

b) students, grades: domain = ${$ Barbara, Pierre, Kateri, Mandeep, Elly $}$

range = ${$ 10, 11, 12 $}$

grades, credits: domain = ${$ 10, 11, 12 $}$

range = ${$ 8, 15, 16, 18, 25 $}$

students, credits: domain = ${$ Barbara, Pierre, Kateri, Mandeep, Elly $}$

range = ${$ 8, 15, 16, 18, 25 $}$

c) The grades-credits relation is NOT a function because 10 in the domain corresponds to 8 and 16 in the range.

For the graph of grades vs credits, If you do vertical line test at $grades = 11$, it touches at two points. Hence, it fails the vertical line test and is therefore not a function.

Thus, the grades-credit relation is not a function.

$$
y=3
$$

$y=3$ is a horizontal line. Use this graph we can conclude for each variable $x$ we gat only one $y=3$, so, this relation is a function

$x=3$

$x=3$ is a vertical line.Looking at this function stuff graphically, what if we had the relation that consists of a set containing just two point $left{(3,-2),(3,2) right}$. We already know that this is not a function, since $x = 3$ goes to each of $y=-2$ and $y=2$

$y=3$ ; horizontal line ; function since it passes vertical-line test

$x=3$ ; vertical line ; NOT a function since it fails vertical line test

i) Substitute $x=0$ into equation $3x+4y=5$, now we get:

$3cdot0+4y=5$

$0+4y=5$ divide both sides by 4

$y=dfrac{5}{4}$

Substitute $x=-2$ into equation $3x+4y=5$, now we get:

$3cdot(-2)+4y=5$

$-6+4y=5$ add 6 on both sides

$-6+4y+6=5+6$

$4y=11$ divide both sides by 4

$y=dfrac{11}{4}$

This is a function for each $x$ we get only one value $y$

ii)Substitute $x=0$ into equation $x^{2}+y^{2}=4$, now we get:

$0^{2}+y^{2}=4$

$y^{2}=4$

$y=pm sqrt{4}$

$y=pm2$

so, if $x=0$, then , $y=-2$ or $y=2$

Substitute $x=-2$ into equation $x^{2}+y^{2}=4$, now we get:

$(-2)^{2}+y^{2}=4$

$4+y^{2}=4$

$y^{2}=0$

$y=0$

This is a not a function because for $x=0$ we get two values of $y$

iii) Substitute $x=0$ into equation $x^{2}+y=2$, now we get:

$0^{2}+y=2$

$0+y=2$

$y=2$

Substitute $x=-2$ into equation $x^{2}+y=2$, now we get:

$(-2)^{2}+y=2$

$4+y=2$

$y=2-4$

$y=-2$

This is a function because for each value $x$ we get only one value of $y$.

iv) Substitute $x=0$ into equation $x+y^{2}=0$, now we get:

$x+y^{2}=0$

$0+y^{2}=0$

$y^{2}=0$

$y=0$

iv) Substitute $x=-2$ into equation $x+y^{2}=0$, now we get:

$-2+y^{2}=0$

$y^{2}=2$

$y=pmsqrt{2}$

so, if $x=-2$, then $y=-sqrt{2}$ or $y=sqrt{2}$

This is a not a function because for $x=-2$ we get two values of $y$

a) i) 1.25 ; 2.75 ii) $pm$ 2 ; 0 iii) 2; -2 iv) 0 ; $pmsqrt{2}$

b) functions: (i) and (iii)

c) From the graph, apply vertical-line test or solve for $y$ in terms of $x$ then see if there are multiple possible values of $x$.

a) $y=sqrt{x+2}$

We know $sqrt{x}$ is defined for $xgeq0$, so, in our case, $y=sqrt{x+2}$ is defined for $x+2geq0$, or $xgeq-2$

For each $xgeq0$ we get only one value of $y$.
We see graph passes the vertical-line test, showing that
for each x-value in the domain there is only one
y-value in the range.

The relation $y=sqrt{x+2}$ is a function.

b) $y=2-x$

For each $x$ we get only one value of $y$.
We see graph passes the vertical-line test, showing that
for each x-value in the domain there is only one
y-value in the range.

The relation $y=2-x$ is a function.

c) $3x^{2}-4y^{2}=12$

At least one vertical line drawn on
the graph intersects the graph at
two points.

This relation $3x^{2}-4y^{2}=12$ not the graph of
a function.

d) $y=-3(x+2)^{2}-4$

For each $x$ we get only one value of $y$.
We see graph passes the vertical-line test, showing that
for each x-value in the domain there is only one
y-value in the range.

The relation $y=-3(x+2)^{2}-4$ is a function.

Numerically, we can solve for $y$ in terms of $x$

$x-y^2=2$

$y^2=x-2$

$y=pmsqrt{x-2}$

For instance if $x=6$

$y=pmsqrt{6-2}=pmsqrt{4}=pm 2$

$y=2$ or $-2$

Each value of $x$ does NOT have a unique value of $y$. Therefore, it is NOT a function.

not a function

graphically, it fails vertical line test

numerically, at $x=6$, $y=pm 2$

a) The domain is the set of possible values of independent variable while the range is the set of all possible values of dependent variable.

The distance traveled by a car cannot be negative so the domain is

${ xinbold{R};|;xgeq 0}$

To find the range, notice that the cost cannot be less than $$44$

$$
{ yinbold{R};|;ygeq 44}
$$

a) domain = ${ xinbold{R};|xgeq 0}$ , range = ${ yinbold{R};|;ygeq 44}$

b) distance can’t be negative, cost cannot lower than minimum daily rental charge

c) function; passes vertical line test

The domain is the set of integers and the range should be less than $5$.

The domain is real numbers less than or equal to 10 and the range must be greater than $-5$.

The point at $x=10$ must be solid while the point at $y=-5$ must be open.

Characteristics of a function:

1) For each value $x$ , function $y=f(x)$ has only one value $y$

(Each element $x$ of the domain has only one
corresponding element $y$ in the range)

2)If any vertical line drawn on the
graph intersects the graph at only
one point

Examples of a function :

1) $y=2x-3$ linear function

2) $y=x^2+3x-4$ quadratic function

3) $3x^{2}+2y-1=0$

4) $y=sqrt{2x+1}$

Not- examples of a function:

1) $x^{2}+y^{2}=9$

2) $3x^{2}-4y^{2}=24$

3) $x+2y^{2}=0$

4) $y=pmsqrt{x+y}$

b) The weight can’t be negative so the domain is

domain = ${ xinbold{R};|;xgeq 0}$

The cost can’t also be negative and there are no minimum fees required so the range is

range = ${ yinbold{R};|;ygeq 0}$

c) The graph can be drawn as follows.

c) In the current system, if the weight is 95 kg you would have to pay $$380$ while you pay only $$350$ for 100 kg. This is unfair.

A recommended system would be:

For items less than $100$ kg, charge $$4$ per kilogram, for items more than $100$ kg, charge $$400$ plus $$3.5$ per kilogram that exceeds $100$ kg. The graph looks like this.

a) Each order corresponds to a single cost.

b) domain = ${ xinbold{R};|;xgeq 0}$ , range = ${ yinbold{R};|;ygeq 0}$

c) see graph inside

d) for $y=4x$ for $x<100$, $y=400+3.5(x-100)$ for $xgeq 100$