Downloaded from – www.vcehelp.com.au Created by VCE student Nicholas Buttigieg for Maths Methods Unit 1/2 Here is an article for all of the functions and graphs used in Maths Methods Units 1 & 2. You can use these as study notes if you like. To look for any diagrams on the shape of graphs, refer to your textbook. Linear Graphs ã The form is y = mx + c, where m is the gradient and c is the yintercept. ã ã ã ã ã ã ã ã ã ã The gradient between two points is equal to m = = Two lines with the same g
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Downloaded from –www.vcehelp.com.au Created by VCE student Nicholas Buttigieg for Maths Methods Unit 1/2Here is an article for all of the functions and graphs used in Maths Methods Units 1 & 2. Youcan use these as study notes if you like. To look for any diagrams on the shape of graphs,refer to your textbook.
Linear Graphs
ã
The form is
y
=
mx
+
c
, where
m
is the gradient and
c
is the
y
intercept.
ã
The gradient between two points is equal to
m
= =
ã
Two lines with the same gradient are parallel to each other and do not intersect.
ã
To find the equation of a straight line given two points, use the formula:
m
(
x

x
1
) =
y

y
1
, where
m
is the gradient.
ã
When given equations of the form
ax
+
by
+
c
= 0, then transpose to
y
=
mx
+
c
ã
Vertical lines are of the form
x
=
a
, where the
x
axis intercept is given as (
a
, 0).
ã
Horizontal lines are of the form
y
=
c
, where the
y
axis intercept is given as (0,
c
).
ã
Perpendicular lines can be calculated using the formula:
m
1
m
2
= 1.
ã
To calculate the tangent of the angle of slope, calculate the gradient, then substituteit into tan to get the angle.
ã
The distance between two points is given as
ã
The midpoint of a line segment is given as
Parabolas
ã
The form is
y
=
ax
2
+
bx
+
c
, which is known as polynomial form.
ã
The form is
y
=
a
(
x – b)
2
+
c
, which is known as turning point form.
ã
Parabolas of the form
y
=
ax
2
, in which the axis of symmetry is
x
= 0, and the turningpoint is (0,0). If
a
is greater than 1, the graph will be narrower than
y
=
x
2
. If it isbetween 0 and 1, the graph will be wider. If
a
is negative, then the graph is invertedor reflected in the
x
axis.
ã
Parabolas of the form
y
=
ax
2
+
c
involve a translation up or down the
y
axis. Theaxis of symmetry is
x
= 0 and the turning point is (0,
c
), which correlates to the
y
intercept.
ã
Parabolas of the form
y
=
a
(
x
–
b
)
2
involve a translation left or right along the
x
axis.The axis of symmetry is
x
= 
b
and the turning point is (
b
, 0), which correlates to the
x
intercept.
ã
The turning point of a parabola is (
b
,
c
) from the turning point form.
ã
The turning point of a parabola from polynomial form can be found by completing thesquare so it is in turning point form.
ã
To find
x
intercepts, either factorise the equation or use the quadratic formula:
x
=
ã
The axis of symmetry is
x
= 
b
from the turning point form.
ã
The axis of symmetry is
x
= from the polynomial form.
ã
The discriminant is equal to
Δ
=
b
2
– 4
ac
.
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ã
If
Δ
< 0, then there are no real solutions.
ã
If
Δ
= 0, then there is one rational solution.
ã
If
Δ
> 0, then there are two solutions. If it results in a perfect square, there are tworational solutions. If it is not a perfect square, there are two irrational solutions.
Rectangular Hyperbolas
ã
The form is
y
=
a
/ (
x

h
) +
k
.
ã
If
a
is greater than 1, then the graph is dilated from the
x
axis, or moves further awayfrom the
x
axis, compared to the graph of
y
= 1 /
x
. If
a
is between 0 and 1, then thegraph is dilated to the
x
axis, or moves closer to the
x
axis. If
a
is negative, then thegraph is inverted or reflected in the
y
axis.
ã
The asymptotes are given as
x
= 
h
and
y
=
k
.
ã
The
x
intercepts can be calculated by substituting
y
= 0.
ã
The
y
intercepts can be calculated by substituting
x
= 0.
The Truncus
ã
The form is
y
=
a
/ (
x
–
h
)
2
+
k
.
ã
If
a
is greater than 1, then the graph is dilated from the
x
axis, or moves further awayfrom the
x
axis, compared to the graph of
y
= 1 /
x
2
. If
a
is between 0 and 1, then thegraph is dilated to the
x
axis, or moves closer to the
x
axis. If
a
is negative, then thegraph is inverted or reflected in the
x
axis.
ã
The asymptotes are given as
x
= 
h
and
y
=
k
.
ã
The
x
intercepts can be calculated by substituting
y
= 0.
ã
The
y
intercepts can be calculated by substituting
x
= 0.
The Square Root Function
ã
The form is
y
=
a
√

(x
–
h
) +
k
ã
If
a
is positive, then the top half of the graph will be sketched. If
a
is negative, thenthe bottom half of the graph will be sketched.
ã
The stationary point, or turning point, is determined by the coordinates (
h
,
k
).
ã
If the expression under the radical is negative, then the graph is reflected in the
y
axis.
ã
If the entire equation is negative, then the graph is reflected in the
x
axis.
ã
The
x
intercepts can be calculated by substituting
y
= 0.
ã
The
y
intercepts can be calculated by substituting
x
= 0.
Circles and Semicircles
ã
The form is
x
2
+
y
2
=
r
2
for circles with a centre at (0,0).
ã
The form is (
x
–
h
)
2
+ (
y
–
k
)
2
=
r
2
for circles not centred at the srcin. This is equal tothe form
x
2
+
y
2
– 2
hx
– 2
ky
+
c
= 0, where
c
=
h
2
+
k
2
–
r
2
.
ã
The centre of the circle is denoted as (
h
,
k
).
ã
The radius of the circle is equal to
r
.
ã
Circles of the form
x
2
+
y
2
– 2
hx
– 2
ky
+
c
= 0 can be sketched by completing thesquare, so it is in the form (
x
–
h
)
2
+ (
y
–
k
)
2
=
r
2
.
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ã
When sketching semicircles, draw the graph of the circle. If there is a positive sign infront of the radical, then draw the top half of the circle. If there is a negative sign,then draw the bottom half of the circle.
ã
The
x
intercepts can be calculated by substituting
y
= 0.
ã
The
y
intercepts can be calculated by substituting
x
= 0.
Cubic Graphs
ã
The form is
y
=
ax
3
+
bx
2
+
cx
+
d
, which is in polynomial or factor form.
ã
The form is
y
=
a
(
x
–
h
)
3
+
k
, which is the basic form.
ã
The point of inflexion from the basic form is given as (
h
,
k
). This only goes for graphs that resemble this form.
ã
Cubic graphs are said to take two distinct patterns: the basic form and the factor form, plus their negative forms.
ã
To find the
x
intercepts, divide the polynomial so it is in factor form. Then use theNull Factor Law to find the
x
intercepts.
ã
If there is a repeated factor in the equation, then the graph will only have two
x
intercepts.
ã
If, when factorised, one linear factor is produced, then there is one
x
intercept. Thisis because the other factors cannot be simplified another further.
ã
Before sketching the graph, draw a sign diagram that helps you to sketch the graphmore accurately.
ã
The
y
intercept can be found by substituting
x
= 0.
Quartic Graphs
ã
The form is
y
=
ax
4
+
bx
3
+
cx
2
+
dx
+
e
, which is in polynomial or factor form.
ã
The form is
y
=
a
(
x
–
h
)
4
+
k
, which is the basic form.
ã
The turning point from the basic form is given as (
h
,
k
). This only goes for graphsthat resemble this form.
ã
Quartic graphs are said to take 6 distinct patterns, which are:
y
=
ax
4

y
=
ax
4
+
cx
2

y
=
ax
2
(
x
–
b
)(
x
–
c
)
y
=
a
(
x
–
b
)
2
(
x
–
c
)
2

y
=
a
(
x
–
b
)(
x
–
c
)
3

y
=
a
(
x
–
b
)(
x
–
c
)(
x
–
d
)(
x
–
e
)
ã
To find the
x
intercepts, divide the polynomial so it is in factor form. Then use theNull Factor Law to find the
x
intercepts.
ã
When factorised, the number of factors is the same as the number of
x
intercepts.
ã
Before sketching the graph, draw a sign diagram that helps you to sketch the graphmore accurately.
ã
The
y
intercept can be found by substituting
x
= 0.
Exponential and Logarithmic Functions
ã
The form is
y
=
k
×
a
bx
+
c
.
ã
For simple graphs of the form
y
=
a
x
, the only asymptote is
y
= 0 or the
x
axis. The
y
intercept will always be equal to 1.
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ã
If the value of
a
is greater than 1, the graph has a positive gradient. This means thatwhen
x
approaches
∞
, then
y
will increase exponentially. If the value of
a
is between0 and 1, then this will produce a negative index. Therefore, this is the same as thegraph with the positive index, but is reflected in the
y
axis.
ã
For graphs of the form
y
=
k
×
a
bx
+
c
, the following apply: The horizontal asymptote is equal to
y
=
c
. The
y
intercept is equal to (0,
k
+
c
) If
b
is a number other than 1, then every
x
value must be divided by
b
to get thepoints of the transformed graph. If
k
is a number other than 1, then every
y
value must be multiplied by
k
to get thepoints of the transformed graph. If
k
is negative, then the graph of its positive form is reflected in the horizontalasymptote.
ã
Logarithmic functions are the inverse of exponential functions. This means that theyare exponential functions reflected in the line
y
=
x
.
ã
For simple graphs of the form
y
= log
a
x
, the only asymptote is
x
= 0 or the
y
axis.The
x
intercept will always be equal to 1.
ã
If the value of
a
is greater than 1, the graph has a positive gradient. This means thatwhen
x
approaches
∞
, then
y
will increase logarithmically. If the value of
a
isbetween 0 and 1, then this will produce a negative logarithm. Therefore, this is thesame as the graph with the positive index, but is reflected in the
x
axis.
ã
For graphs of the form
y
= log
a
(
x
–
c
), the following apply: The vertical asymptote is equal to
x
=
c
. The
x
intercept is equal to (1 +
c
, 0).
Sine and Cosine Functions
ã
The form is
y
=
a
sin
n
(
x
±
b
) +
c
or
y
=
a
cos
n
(
x
±
b
) +
c
.
ã
The basic graph is
y
= sin
x
. It has the following characteristics: The graph has a wavelength or period of 2
π
units. It is a periodic function. The maximum and minimum values are
y
= 1 and
y
= 1. This means that theamplitude is equal to 1. The
y
intercept is equal to (0, 0).
ã
The other basic graph is
y
= cos
x
. It has the following characteristics: The graph has a wavelength or period of 2
π
units. It is a periodic function. The maximum and minimum values are
y
= 1 and
y
= 1. This means that theamplitude is equal to 1. The
y
intercept is equal to (0, 1).
ã
In the graphs of the form
y
=
a
sin (
nx
) and
y
=
a
cos (
nx
), the followingcharacteristics apply: The graph has a wavelength or period of 2
π
/
n
. The amplitude is always equal to
a
. The maximal domain is equal to
R
. The range of each function is [
a
,
a
]. If
a
is negative, then the graph of the positive form is reflected in the
x
axis.
ã
In graphs of the form
y
=
a
sin
n
(
x
±
b
) and
y
=
a
cos
n
(
x
±
b
), a horizontaltranslation along the
t
axis occurs, with the graph moving
b
units to the left or right.