Describe the Graph of Y Mx Where M 0
Describe the Graph of Y Mx Where M 0
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This topic is relevant for:
y=mx+c
Here we will learn about
y = mx + c
including how to recognise the gradient and
and rearrange an equation into the standard grade of
At that place are besides
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
is the
slope
of the line (how steep the line is) and
c
is the
(the point in which the line crosses the
is a
linear
equation and the variables
and
relate to coordinates on the line.
When nosotros input a value for
into the equation
This means that
is an
independent variable, and
is a
dependent variable
every bit it is determined by the value of
E.g.
Let’s look at the line
This graph has a
gradient
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
quadratic - If the variables are raised to a power no higher than
3
nosotros refer to it equally a
cubic, and so on.
What is y=mx+c?
The gradient ‘one thousand’
The gradient of the line tells usa
how steep the line is.
We utilize the alphabetic character
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
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
is the indicate of
intersection
between the straight line
when
Eastward.m.
Let’south look at the equation
To find the
intercept we substitute
into the equation.
\brainstorm{aligned} &y=(five \times 0)+7\\\\ &y=7 \end{aligned}
So when
giving united states the coordinate
This is the
Here is a quick summary of some equations in the form
with the gradient and
intercept highlighted.
gradient
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
-
Rearrange the equation to make
y
the subject -
Substitute
x = 0
into the equation to find the
y -intercept -
Land the coefficient of
ten
(the gradient)
How to use y=mx+c
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y = mx + c examples
Example 1: y=mx+c course
State the slope and
-
Rearrange the equation to make
y
the subject field
The equation
is already in the general form of
so we can progress to pace
straight away.
ii
Substitute
into the equation to find the
When
\begin{aligned} &y=(-three\times{0})+eight\\\\ &y=0+8\\\\ &y=viii \end{aligned}
The
is
three
State the coefficient of
(the gradient)
The coefficient of
is
The gradient of the line
is
Solution:
Instance ii: y=mx+c form
Land the gradient and
Rearrange the equation to make
the subject
Here we take the two terms of
and
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
.
\begin{aligned} &y=7-x\\\\ &y=-x+seven \end{aligned}
Substitute
into the equation to discover the
\brainstorm{aligned} &y=-0+7\\\\ &y=0+seven\\\\ &y=7 \finish{aligned}
The
is
State the coefficient of
(the slope)
The coefficient of
is
as
is the same as
The gradient of the line
is
Solution:
Example 3: rearrange the equation when ten is the subject
State the gradient and
Rearrange the equation to make
the discipline
Here we need to make
the subject field of the equation. We therefore need to make sure we isolate
on one side of the equals sign so that we express the equation in the form
Substitute
into the equation to find the
\begin{aligned} &y=0-10\\\\ &y=-ten \end{aligned}
The
is
Land the coefficient of
(the gradient)
The coefficient of
is
every bit
is the same equally
The gradient of the line
is
Example 4: changing the subject including a fraction
Country the gradient and
Rearrange the equation to make
the subject
Here we demand to make
the subject area of the equation. We therefore need to make sure we isolate
on one side of the equals sign so that we limited the equation in the form
Substitute
into the equation to discover the
\begin{aligned} &y=(\frac{one}{3}\times{0})+\frac{v}{2}\\\\ &y=0+\frac{five}{two}\\\\ &y=\frac{v}{two} \end{aligned}
The
-intercept of the line
is
\[\frac{5}{2}, \quad \text{or} \quad c=\frac{5}{ii}.\]
Country the coefficient of
(the gradient)
\[\frac{1}{3}.\]
The gradient of the line
is
\[\frac{1}{3}, \quad \text{or} \quad m=\frac{1}{iii}.\]
Example 5: changing the subject area including brackets
Land the gradient and
Rearrange the equation to make
the bailiwick
Hither we need to make
the subject of the equation. We therefore need to make sure we isolate
on i side of the equals sign and then that nosotros express the equation in the form
Substitute
into the equation to discover the
\begin{aligned} &y=(\frac{iii}{4}\times{0})+v\\\\ &y=0+5\\\\ &y=5 \terminate{aligned}
The
is
State the coefficient of
(the gradient)
\[\frac{three}{four}.\]
The gradient of the line
is
\[\frac{3}{4}, \quad \text{or} \quad thousand=\frac{3}{4}.\]
Example six: irresolute the subject field including decimals
State the gradient and
\[10=\frac{y+0.85}{0.2}.\]
Rearrange the equation to brand
the subject
Hither we need to brand
the subject of the equation. We therefore need to make sure we isolate
on one side of the equals sign so that we express the equation in the class
Substitute
into the equation to discover the
\begin{aligned} &y=(0.2\times{0})-0.85\\\\ &y=0-0.85\\\\ &y=-0.85 \finish{aligned}
The
-intercept of the line
\[x=\frac{y+0.85}{0.ii}\]
is
State the coefficient of
(the gradient)
The coefficient of
is
\[10=\frac{y+0.85}{0.ii}\]
is
or
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.
is rearranged to make
✘
-
Stating the value of
chiliad
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
.
Take case
in a higher place when
. The gradient would be stated as
as this is the coefficient of
however the value of the
-intercept would exist incorrectly stated equally
The correct respond for the
intercept is
as we can rearrange the equation to make
the discipline. Come across below.
-
Mixing up the gradient and the y-intercept
Have, for example, the equation
is
and the value of
when
is
and the
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
chiliad=\frac{ii}{3}
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)
Prove answer
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)
Prove answer
(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)
Bear witness answer
(3)
Learning checklist
You accept now learned how to:
- reduce a given linear equation in 2 variables to the standard course y=mx+c
Still stuck?
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Describe the Graph of Y Mx Where M 0
Source: https://thirdspacelearning.com/gcse-maths/algebra/y-mx-c/