# Describe the Graph of Y Mx Where M 0

Describe the Graph of Y Mx Where M 0

## y=mx+c

Here we will learn about

y = mx + c

including how to recognise the gradient and
y-intercept from the equation

y = mx + c

and rearrange an equation into the standard grade of

y = mx + c
.

At that place are besides

y = mx + c

worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go side by side if you’re yet stuck.

## What is y=mx+c?

The equation

y = mx + c

is the general equation of any direct line where

m

is the
slope
of the line (how steep the line is) and

c

is the
y-intercept
(the point in which the line crosses the
y-centrality).

y = mx + c
is a
linear
equation and the variables
10
and
y
relate to coordinates on the line.
When nosotros input a value for
10
into the equation

y = mx + c
, we get a upshot for
y.
This means that
ten
is an
independent variable
, and
y
is a
dependent variable
every bit it is determined by the value of
ten.

E.g.

Let’s look at the line
y = 2x + i.

This graph has a

of

2

and a y-intercept of

1
, the coordinate (0,ane).

• The term
linear
is used to describe a straight line where the variables are raised to a power no higher than
ane
• If the variables are raised to a power no higher than
2
we refer to it every bit a
• If the variables are raised to a power no higher than
3
nosotros refer to it equally a
cubic, and so on.

### The gradient ‘one thousand’

The gradient of the line tells usa
how steep the line is
.

We utilize the alphabetic character

grand

to denote the gradient.

Imagine climbing a ladder. If the ladder is really close to the wall, the gradient of the ladder is really steep (you would near be climbing vertically). Taking the base of the ladder away from the wall means that the gradient of the gradient decreases reaching a lower point on the wall.

The bigger the gradient the steeper the line.

E.thousand.
A slope of
2
will generate a line that is much steeper than the gradient of

{\frac{1}{3}}
.

Meet the diagram below.

The slope of a line can exist many different types of number, i.e. fractions, decimals, negatives etc.

### The y-intercept ‘c’

The

y-intercept
is the indicate of
intersection
between the straight line

y = mx + c
, and the
y-axis. The
y-intercept is the value of
y
when
x = 0.

Eastward.m.
Let’south look at the equation
y = 5x + vii.

To find the
y
intercept we substitute
x = 0
into the equation.

\brainstorm{aligned} &y=(five \times 0)+7\\\\ &y=7 \end{aligned}

So when
ten = 0, y = vii,
giving united states the coordinate
(0,vii).

This is the
y-intercept.

Here is a quick summary of some equations in the form

y = mx + c

with the gradient and
y
intercept highlighted.

thou

2

half-dozen

-4

-1

y
-intercept

c

4

-3

0

(the origin)

vi

## How to use y=mx+c

In club to state the gradient and
y-intercept of a line:

1. Rearrange the equation to make
y
the subject
2. Substitute
x = 0
into the equation to find the
y-intercept
3. Land the coefficient of
ten

## y = mx + c examples

### Example 1: y=mx+c course

State the slope and
y-intercept of the line
y = −3x + 8.

1. Rearrange the equation to make
y
the subject field

The equation
y = −3x + 8
is already in the general form of

y=mx+c

so we can progress to pace
2
straight away.

ii
Substitute
x = 0
into the equation to find the
y-intercept

When
x = 0,

\begin{aligned} &y=(-three\times{0})+eight\\\\ &y=0+8\\\\ &y=viii \end{aligned}

The
y-intercept of the line
y = −3x + 8
is
eight, or
c = 8.

three
State the coefficient of
x

The coefficient of
x
is
−iii.

The gradient of the line
y = −3x + eight
is
−iii, or
g = −3.

Solution:
grand = −iii, c = 8

### Instance ii: y=mx+c form

Land the gradient and
y-intercept of the line
y = 7 − x.

Here we take the two terms of

+7

and

−x

and then we need to change the guild they appear on the correct hand side of the equals sign so that we get the equation in the form

y=mx+c

.

\begin{aligned} &y=7-x\\\\ &y=-x+seven \end{aligned}

\brainstorm{aligned} &y=-0+7\\\\ &y=0+seven\\\\ &y=7 \finish{aligned}

The
y-intercept of the line
y = seven − 10
is
vii, or
c = 7.

The coefficient of
10
is
−1
as
−x
is the same as
−1x.

The gradient of the line
y = vii − x
is
−i, or
thousand = −1.

Solution:
m = −1, c = 7.

### Example 3: rearrange the equation when ten is the subject

State the gradient and
y-intercept of the line
10 = y + x.

Here we need to make
y
the subject field of the equation. We therefore need to make sure we isolate
y
on one side of the equals sign so that we express the equation in the form

y=mx+c
.

\begin{aligned} &y=0-10\\\\ &y=-ten \end{aligned}

The
y-intercept of the line
ten = y + 10
is
−x, or
c = −10.

### Example 4: changing the subject including a fraction

Country the gradient and
y-intercept of the line
2x = 6y − fifteen.

Here we demand to make
y
the subject area of the equation. We therefore need to make sure we isolate
y
on one side of the equals sign so that we limited the equation in the form

y=mx+c
.

\begin{aligned} &y=(\frac{one}{3}\times{0})+\frac{v}{2}\\\\ &y=0+\frac{five}{two}\\\\ &y=\frac{v}{two} \end{aligned}

The

y

-intercept of the line

2x = 6y −
fifteen

is

$\frac{5}{2}, \quad \text{or} \quad c=\frac{5}{ii}.$

$\frac{1}{3}.$

The gradient of the line

2x = 6y − 15

is

$\frac{1}{3}, \quad \text{or} \quad m=\frac{1}{iii}.$

### Example 5: changing the subject area including brackets

Land the gradient and
y-intercept of the line
3x = iv(y − five).

Hither we need to make
y
the subject of the equation. We therefore need to make sure we isolate
y
on i side of the equals sign and then that nosotros express the equation in the form

y=mx+c
.

\begin{aligned} &y=(\frac{iii}{4}\times{0})+v\\\\ &y=0+5\\\\ &y=5 \terminate{aligned}

The
y-intercept of the line
3x = (y −
5)
is
5, or
c = v.

$\frac{three}{four}.$

The gradient of the line

3x = (y − 5)

is

$\frac{3}{4}, \quad \text{or} \quad thousand=\frac{3}{4}.$

### Example six: irresolute the subject field including decimals

State the gradient and
y-intercept of the line

$10=\frac{y+0.85}{0.2}.$

Hither we need to brand
y
the subject of the equation. We therefore need to make sure we isolate
y
on one side of the equals sign so that we express the equation in the class

y=mx+c

.

\begin{aligned} &y=(0.2\times{0})-0.85\\\\ &y=0-0.85\\\\ &y=-0.85 \finish{aligned}

The

y

-intercept of the line

$x=\frac{y+0.85}{0.ii}$

is
−0.85, or
c = −0.85.

The coefficient of
x
is
0.2.

$10=\frac{y+0.85}{0.ii}$

is
0.2,
or
grand = 0.2.

### Common misconceptions

• Incorrect inverse operation

When rearranging equations, instead of applying the inverse operation to the value being moved, the value is merely moved to the other side of the equals sign.

E.g.

y + 5 = 2x

is rearranged to make

y = 2x + 5

• Stating the value of
and
c

A common error is to incorrectly state the values of m and c as a consequence of not rearranging the equation and then that it is in grade of

y=mx+c

.

Take case

3

in a higher place when

x = y + 10

. The gradient would be stated as

1

as this is the coefficient of

10

however the value of the

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y

-intercept would exist incorrectly stated equally

10.

The correct respond for the

y

intercept is

−10

as we can rearrange the equation to make

y

the discipline. Come across below.

• Mixing up the gradient and the y-intercept

Have, for example, the equation
y = 10 + 3x. Here, the coefficient of
x
is
3
and the value of
y
when
x = 0
is
ten. This means that the slope is
three
and the
y-intercept is
10, not the other way around.

### Exercise y=mx+c questions

The coefficient of

x

is

-5
, so

thou=-5

When

x=0, y=9,

so
c=9.

y=6-x

y=-x+vi

The coefficient of

x

is

-1
, and so

m=-1

When

10=0, y=six,

so
c=half dozen.

m=\frac{1}{two}, \; c=-\frac{5}{2}

y=\frac{one}{2}ten-\frac{v}{2}

The coefficient of

x

is

\frac{ane}{two}
, so

m=\frac{1}{two}

When

10=0, y=-\frac{v}{2},

so

c=-\frac{5}{2}.

m=\frac{three}{5}, \; c=\frac{6}{5}

y=\frac{3}{v}x+\frac{6}{5}

The coefficient of

x

is

\frac{iii}{5}
, so

m=\frac{3}{v}

When

x=0, y=\frac{6}{5},

then

c=\frac{half dozen}{5}.

y=\frac{two}{three}x-three

The coefficient of

10

is

\frac{two}{iii}
, so

When

ten=0, y=-3

so

c=-three.

m=-\frac{2}{3}, \; c=\frac{1}{iii}

y=-\frac{2}{3}x+\frac{1}{iii}

The coefficient of

ten

is

-\frac{2}{iii}
, so

k=-\frac{2}{3}

When

x=0, y=\frac{1}{3},

so

c=\frac{1}{3}.

### y=mx+c GCSE questions

1.
Given that the coordinate

(three,4)

lies on the line

y=3x+c

calculate the

y
-intercept of the directly line.

(2 Marks)

Substitute

x=3
and

y=4
into

y=3x+c
to get

4=(three\times3)+c

(1mark)

(i)

4=nine+c
so

c=-5

(1mark)

(1)

2.  (a)  The coordinate

A=(0,ii)

lies on a straight line. The gradient of the line is

5
. Using this data, state the equation of the straight line.

(b)
Write the equation of a line that is parallel to a) in the form

y=mx+c
.

(4 Marks)

(a)

A

is the

y
-intercept then

c=2

or when

x=0, y=2
so

c=2

(ane)

y=5x+c

(1mark)

(1)

y=5x+2

(1mark)

(1)

(b)

y=5x+c
where

c

two

(1mark)

(1)

3.  Show

m=2

for the direct line

8x-4y=12.

(3 Marks)

(3)

## Learning checklist

You accept now learned how to:

• reduce a given linear equation in 2 variables to the standard course y=mx+c

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## Describe the Graph of Y Mx Where M 0

Source: https://thirdspacelearning.com/gcse-maths/algebra/y-mx-c/