Formula Sheet for Mid-semester Exam
Week 1
Chapter 1.2 – Graphs and Lines
General form of a line:
𝑦 = 𝑚𝑥 + 𝑏
𝑟𝑖𝑠𝑒
 “m” is the gradient of the line (𝑚 = 𝑟𝑢𝑛 )


Where “b” is the y intercept
−𝑏
is x intercept
𝑚
Point Gradient Formula
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
 (𝑥1 , 𝑦1 ) is a point on the line
𝑟𝑖𝑠𝑒
 𝑚 is the gradient of the line (𝑚 = 𝑟𝑢𝑛 )
Chapter 2.1 – Functions
Vertical transformation
Where 𝑓(𝑥) = 𝑓(𝑥) − 𝑘
The function is shifted downwards by “k” units
Horizontal transformation
Where 𝑓(𝑥) = 𝑓(𝑥 − 𝑘)
The function is shifted to the right by “k” units
Stretching/squeezing the function
Where 𝑓(𝑥) = 2𝑓(𝑥)
The function rises/falls double as quickly
𝑓(𝑥)
As such, where 𝑓(𝑥) = 2 , the function rises/falls half as quickly
Reflection
Where f (x) = - f (x), the graph is reflected about the x axis
Where x and y are inverted (i.e. x = y 2 ) the graph is perpendicular to how the graph will have been
Shapes of graphs:
𝑓(𝑥) = 𝑚𝑥 + 𝑏  line
f (x) = x 2  parabola
f (x) = x 3  hyperbola
f (x) = x  “v” shaped graph
Chapter 2.3 – Quadratic Functions
The Quadratic Function
f (x) = ax 2 + bx + c  General form
f (x) = (x ± d)(x ± e)  Intercept form
f (x) = (x ± f )2  vertex form
To find the x-intercepts of a quadratic function:
Either: factorize to intercept form, then equate each bracket to zero
Or: use the quadratic formula:
-b ± b2 - 4ac
2a
To find the vertex form of a quadratic function:
Either: use the quadratic formula to find the x-intercepts, and use the midpoint between the two
intercepts:
Or: find where f '(x) = 0
Or: complete the square
e.g. f (x) = -x 2 +8x - 9
-b
1. take “ ( )2 ", and add/subtract it to the formula
2
-b
-8
a. f (x) = -x 2 +8x +16 -16 - 9 ( )2 = ( )2 =16
2
2
2
b. f (x) = -x +8x +16 -16 - 9
c. = -1(x 2 -8x +16)- 9 +16
2. Find numbers that multiplies to give you “ (
of the equation.
a. Multiplies for 16, sums to -8 = -4
b. = -1(x - 4)(x - 4)+ 7
-b 2
) ” and adds to give you “b”: these are the roots
2
c. = -(x - 4)2 + 7
3. Analyze the transformations:
a. Here, we have a parabola, inverted, shifted upwards by 7 units and across 4 units to the
right.
Week 2
Chapter 2.4 – Polynomials and rational function
Degree of a polynomial
Is the highest “power” in the polynomial chain:
e.g. f (x) = 8x 20 + 6x 2 + 2x +1, degree of the polynomial is “20”
Y-intercepts
Where x=0
X-Intercepts
Where y=0
Note, where the polynomial has been factorized, equating each bracket to zero will give the xintercepts (/roots)
e.g. f (x) = (2 - 5x)(x - 6)(x -1)
-2
Roots are: (
,0)(6,0)(1,0)
5
Finding asymptotes
n(x)
Where f (x) =
d(x)
Vertical Asymptotes:
1. After cancelling common factors, where d(x) = 0 , there is a vertical asymptote
Horizontal Asymptotes
1. If the degree of n(x) < the degree of d(x) , y = 0 is the horizontal asymptote
2. If the degree of n(x) = the degree of d(x) , y = a is the horizontal asymptote:
b
a. a is the leading coefficient of n(x)
b. b is the leading coefficient of d(x)
3. If degree of n(x) > degree of d(x) , there is no horizontal asymptote
Chapter 2.5 – Exponential functions
Note: graphs shift as per regular functions, see Chapter 2.1 – Functions.
A graph has an exponential shape where f (x) = b x , x ¹1, x > 0 .
Properties of f (x) = b x
1. All graphs go through
for any base b
2. The graph is continuous
3. The x axis is a horizontal asymptote
4. Where b >1, b x increases as x increases
5. Where 0 < b <1, b x decreases as x increases
Exponent Laws
Where a and b are positive, a ¹1, b ¹1 , x & y are real.
1. a x a y = a x+y
ax
2.
= a x-y
y
a
3. (a x ) y = a xy
Further:
1. a x = a y iff x = y
4. (ab)x = a x b x
a
ax
( )x = x
b
5. b
2. where x ¹ 0, a x = b x iff a = b
Common bases
10 & e
Note, growth/decay formulae are often in the form y = cekt , c & k are constants, t is time.
Chapter 2.6 – Logarithmic Functions
y = loga x iff x = a y
that is: a = bc : logb a = c
Log properties
Where b, M and N are positive, b ¹1and p & x are real numbers
1. logb 1 = 0
M
6. logb
= logb M - logb N
2. logb b =1
N
7. logb M p = plogb M
3. logb b x = x
8. logb M = logb N iff M = N
4. blogb x = x, x > 0
5. logb MN = logb M + logb N
Changing base of log etc.
Note, calculator has "log" = log10 and "ln" = loge
ln x

= log b x
ln b

e x ln b = b x
Chapter 3.1 – simple interest
I = P· R·t , where I is interest, P is principle, r is the annual simple interest rate, and t is the time in
years.
Note: do not forget to add the principle again when working out future value, since this formula only
works out interest
Chapter 3.1 – compound interest
The compound interest formula
r
A = P(1+ i)n = P(1+ )mt : where A is the future value at the end of n periods, P is the principle, r is the
m
annual the annual nominal rate of interest, m is the amount of compounding periods per year, i is the
interest rate per compounding period, n is the total number of compounding periods.
Note: make sure “i” and “n” are in the same units of time.
Continuous compound interest
A = Pert , r is the annual compounding rate, t is time in years.
Computing growth time
Since A = P(1+ i)n , ln A = n ln(P(1+i)).
Annual percentage yield
r
APY = (1+ )m -1, or, if compounded continuously, APY = er -1
m
Chapter 3.4 – Annuities
Strategy (make into flowchart for final notes?):
1. Make a timeline of payments
2. If single payment: either simple or compound interest
3. If multiple:
a. Payments into an account increasing in value (FV)
b. Payments being made out of an account decreasing in value (PV)
c. All amortization is PV.
Future value of an ordinary annuity
(1+ i)n -1
FV = PMT (
) , where FV is the future value, PMT is the periodic payment, i is the rate per
i
period, n is the number of periods/payments.
Present value of an ordinary annuity
1- (1+ i)-n
PV = PMT(
)
i
Week 3
Chapter 10.4 – the derivative
Slope of a secant between two points
f (a) - f (b)
a-b
Average rate of change (slope of a secant between x and x+h)
f (a + h) - f (a)
,h ¹ 0
h
The derivative from first principles
f (x + h) - f (x)
note: it is most probable that the h on the denominator will go
lim
h®0
h
This will fail if the line is non-differentiable at a point, e.gg where the graph:
 is not continuous
 has a sharp corner
 has a vertical tangent
Chapter 10.5 – basic differentiation properties
1. Constant
f (x) = c ® f '(x) = 0
2. Just an x
f (x) = x ® f '(x) =1
3. A power of x
f (x) = x n ® f '(x) = nx n-1
4. A constant*a function
f (x) = k ·u(x) ® f '(x) = k ·u'(x)
5. Sum/difference
f (x) = u(x)± v(x) ® f '(x) = u'(x)± v'(x)
Chapter 11.2 – derivatives of logarithmic and exponential functions
1. Base e exponential
f (x) = e x ® f '(x) = e x
2. Base e exponential with constant in power
3. Other exponential
4. Natural log
5. Other log
f (x) = ecx ® f '(x) = cecx
f (x) = b x ® f '(x) = b x lnb
1
f (x) = ln x ® f '(x) =
x
1 1
f (x) = logb x ® f '(x) =
·
ln b x
Chapter 11.3 – product/quotient rule
The product rule
f (x) = F(x)S(x) ® f '(x) = F(x)S'(x)+ S(x)F '(x)
The quotient rule
T(x) B(x)T '(x) - T(x)B'(x)
f (x) =
®
B(x)
[B(x)]2
Chapter 11.4 – the chain rule
m(x) = f (g(x)) ® m'(x) = f '(g(x))·g'(x)
easy to do via substitution
the above formula means: derivative of whole thing times derivative of bracket
The general derivative rules
d
1.
[ f (x)]n = n[ f (x)]n-1 · f (x)
dx
d
1
2.
ln[ f (x)] =
· f '(x)
dx
f (x)
d f ( x)
3.
e = e f ( x ) · f '(x)
dx
Week 4
Note, in this chapter, sign charts make stuff easier
Chapter 12.1 – first derivatives and graphs
The first derivative gives the slope of a graph at a point: a positive first derivative will give an upward
slope, and vice versa. A “0” or undefined first derivative gives a partition value. It is also a critical
value when it appears in both the domain of f’(x) and f(x), e.g. an asymptote is a partition value but
not a critical value.
Local extrema
Where the first derivative is 0, and the sign of the first derivative changes around it, it is a local
extrema:
1. – 0 +  minimum
2. + 0 -  maximum
3. – 0 – or + 0 +  not a local extrema
Note, where f '(c) = 0 , finding f ''(c) can also identify whether it is a local extrema: where
f '(c) = 0 & f ''(c) = + , it is a local minimum; where f '(c) = 0 & f ''(c) = - , it is a local maximum. This test
is invalid where f ''(c) = 0 .
Chapter 12.2 – second derivatives and graphs
The second derivative describes the concavity of a graph (where f ''(x) > 0 , the concavity is positive
, and f '(x) (/the slope) is increasing; where f ''(x) < 0 , the concavity is negative
and f '(x)
(/the slope) is decreasing.
Point of inflexion
A point of inflexion is where the concavity of the graph changes (and, as such, the sign of the
derivative, too, will change. This occurs where f ''(x) = 0 (or, if it is a vertical point of inflexion,
undefined) the line is continuous and the sign of the second derivative changes about that point.
Graph sketching
1. Analyze f (x) , find domain and intercepts
2. Analyze f '(x) , find partition numbers and critical values and construct a sign chart (to find
increasing/decreasing segments and local extrema)
3. Analyze f ''(x) , find partition numbers and construct a sign chart (to find concave up and down
segments and to find inflexion points)
4. Sketch f (x) : locate intercepts, maxima and minima and inflexion points: if still in doubt, sub
points into f '(x)
Chapter 12.6 – optimization
1. Introduce variables, look for relationships among variables, and construct a mathematical
model of the form Maximize/minimize f (x) on the interval I.
2. Find critical values of f (x) .
3. Find absolute maxima/minima: this will occur at a critical value or at an endpoint of an interval
a. Check that the function is continuous over an interval
b. Evaluate f (x) at the endpoints of the interval
c. Find the critical values of f '(x)
d. The absolute maximum is the largest value found in step “b” or “c”.
Chapter 4.1 – Systems of linear equations in two variables
Simultaneous equations of two lines: isolate a variable and substitute.
Week 5
Chapter 13.1 – antiderivatives and indefinite integrals
Antiderivative is symbolized by F(x) , and may be accompanied by any constant F(x) + C .
Indefinite integral
ò f (x)dx = F(x) + c is a family of antiderivatives
Indefinite integrals of basic functions
1. x to the power of n
2. e to the power of x
x n+1
ò x dx = n +1 + C
ò e x dx = ex + C
n
3. x as a denominator
1
ò x dx = ln x + C, x ¹ 0
Indefinite integrals of a constant multiplied by a function, or, two functions
1.
k · f (x)dx = k · f (x)dx
[ f (x)± g(x)]dx =
2.
ò
ò
ò
ò f (x)dx + ò g(x)dx
Chapter 13.2 – integration by substitution
d
Based on the chain rule:
f [g(x)] = f '[g(x)]· g'(x) (derivative of outside function multiplied by the
dx
derivative of the inside function)
Thus f '[g(x)]·g'(x)dx = f [g(x)]+ C
ò
General indefinite integral formulae
[ f (x)]n+1
1. ò [ f (x)]n · f '(x)dx =
+ C, n ¹ 0
n +1
2.
e f (x) · f '(x)dx = e f ( x) + C
3.
ò
ò
1
· f '(x)dx = ln f (x) + C
f (x)
Integration by substitution
Sometimes it is hard to recognize the form of the function to be integrated (that is: to see which of the
above formulae apply to it). So, we substitute the messy part for “u” and integrate with respect to “u”,
rather than x.
Where y = f (x) ® dy = f '(x)dx
General indefinite integral formulae
un+1
1. ò un du =
+ C, n ¹ -1
n +1
2.
eu du = eu + C
3.
ò
1
ò u du = ln u + C
Method of integration by substitution
1. Select a substitution to simplify the integrand: one such that u and du (the derivative of u) are
present
2. Express the integrand in terms of u and du, completely eliminating x and dx
3. Evaluate the new integral
4. Re-substitute from u to x.
Note, if this is incomplete (i.e. du is not present) you may multiply by the constant factor and divide,
1
1 4
1
1
outside of the integral, by its inverse: e.g. ò
dx = ò
dx = ò
4 dx
4x + 7
4x + 7 4
4 4x + 7
Where integral of u is 1, du=dx. (and find x as “u-k”)
Chapter 13.4 – the definite integral
ò
b
a
f (x)dx is the definite integral of f (x) from x=a to x=b. Worked out by subtracting where x=a from
where x=b. Note, these are not absolute values: above x axis is positive, below is negative: opposite if
other direction.
Properties of a definite integral
1.
2.
3.
4.
5.
ò
ò
ò
ò
ò
a
a
b
a
b
a
b
a
b
a
f (x)dx = 0
f (x)dx = - ò f (x)dx
a
b
k · f (x)dx = k · ò f (x)dx , where k is a constant
b
[ f (x)± g(x)] =
f (x)dx =
ò
c
a
ò
a
b
a
f (x)dx ± ò g(x)dx
b
a
f (x)dx + ò f (x)dx
b
c
Error Bounds
Where f(x) is above the x-axis: f (b) - f (a) ·
b-a
n
The fundamental theorem of calculus
ò
b
a
f (x)dx = F(b) - F(a)
Average value of a continuous function over a period
b
1
f (x)dx
ò
b-a a
Week 6
Chapter 15.1 – functions of several variables
Substitute (x,y,z,…,et.) into the equation given.
Find the shape of the graph by looking at cross sections (e.g. y=0, y=1, x=0, x=1).
Chapter 15.2 – partial derivatives
Derivatives with respect to a certain symbol: watch for signs, all other symbols count as constants
fx (x, y) derived with respect to x ||| fxy (x, y)  derived first with respect to x, then y
Chapter 15.3 – maxima and minima
1. Express the function as z = f (x, y)
2. Find fx (x, y) & fy (x, y) , and simultaneously equate them to find critical values
3. Find f xx (a, b), fxy (a, b)& fyy (a, b) (A, B, and C, respectively)
4. Find A, and AC - B2 .
a. IF AC-B*B>0 & A<0, f(a,b,) is local maximum
b. IF AC-B*B>0 & A>0, f(a,b) is local minimum
c. IF AC-B*B<0, f(a,b) is a saddle point
d. IF AC-B*B=0, test fails
Chapter 15.4 – maxima and minima using Lagrange multipliers
1. Write problem in form
a. max/ min ® z = f (x, y)
b. g(x, y) = 0
2. Form the function F(x, y, l ) = f (x, y)+ l g(x, y)
3. Derive with respect to x, y and lambda
4. Simultaneously equate answers
5. If more than 1 answer, find z values and deduce which is max/min.