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Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 1 linear
relations
Chapter 1 linear
relations
Chapter 1 linear
relations
If I want to… and
recommended ?’s
I will use…
Get the slope and y
intercept from a
linear equation
Calculate slope (m)
y = mx + b
Find the point of
intersection of two
lines graphically
Tables of values
m
y2  y1
x2  x1
Chapter review
p.121, #3 and 4
Find the point of
intersection using
substitution
Chapter review
p.128, #16 & 17
Chapter review test
p.137, #2
m is the slope
b is the y-intercept
Pick your point with the highest y value as point 2
Substitute the x and y values of both points and solve for m
Equation should be in y = mx + b form
If not a word problem and the slope is an integer, use x values -2, -1, 0, 1, 2
(ex. y = -2x + 7… -2 is an integer)
If not a word problem and slope is a fraction, use x values that are
multiples of the denominator of the slope (ex. y 
1
x  3 … ½ is an integer,
2
so you would use x values -4, -2, 0, 2, 4)
Create the tables of values
Plot the tables of values on the graph
Read the point of intersection off the graph
You can verify the point of intersection by substituting the point of
intersection into both equations – for the solution to be correct, the left
side has to equal the right side
Chapter review test
p.137, #1
Chapter 1 linear
relations
Strategy
Substitution method
and algebra
At least one of the two equations must be expressed as y in terms of x, or
x in terms of y, so it can be substituted into the other equation
Ex. y = 3x – 9 and 5x + y = 6, you can plug the “3x – 9” into 5x + y = 6, so you
get 5x + 3x – 9 = 6,… then 8x – 9 = 6,… then 8x = 15, finally x =
15
8
Once you find one of the variables, substitute the answer back into any of
the original equations
Ex. y = 3(
15
45
5
3
) - 9, … then y =
- 9,… then y = 5 – 9,… then y = - 3
8
8
8
8
You can verify the point of intersection by substituting the point of
intersection into both equations – for the solution to be correct, the left side
has to equal the right side
1
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 1 linear
relations
If I want to… and
recommended ?’s
Find the point of
intersection using
elimination
I will use…
Elimination method and
algebra
Chapter review
p.131#21, 22
Chapter review test
p.137, #3
Chapter 1 linear
relations
Draw a graph
representing a
system of equations
with no solution, or
one solution, or an
infinite number of
solutions
Ch. review p.123,
#6 & Ch. review
test p.137, #9
Memorization
Strategy
The first thing you have to do is line up the 2 equations so that the x’s, y’s
and numbers line up, 1 equation above the other, expressing each in the
ax + by = c format… number the top equation (1) & the bottom equation (2)
Then, you see what you have to multiply one or both equations by to end up
with the opposite coefficients in front of two of the same variables
Ex. 3x + 2y = 13 and
-2x + 4y = 2 (these two equations are already lined up)
Ex. 3x + 2y = 13 (1) multiply equation (1) by -2 to get equation (3)
-2x + 4y = 2 (2)
----------------------6x – 4y = -26 (3) (label this equation (3))
-2x + 4y = 2 (2) (copy down this equation again from above)
----------------------8x = -24 (add the two equations and the y’s cancel out)
x=3
(divide both sides by -8 to get the final x)
Once you find one of the variables, substitute the answer back into any of
the original equations
Ex. 3x + 2y = 13,… then 3(3) + 2y = 13,… then 9 + 2y = 13,… then 2y = 4,…
then y = 2
So, the point of intersection is (3,2)
You can verify the point of intersection by substituting the point of
intersection into both equations – for the solution to be correct, the left
side has to equal the right side
No solution is two parallel lines
One solution is two intersecting lines
An infinite number of solutions is two lines drawn on top of each other
2
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 1 linear
relations
If I want to… and
recommended ?’s
Set-up a word
problem
Chapter review
p.121, #5,
p.123, #7-10
p.125, #12-14
p.128, #18-20
p.131, #23-25
I will use…
Let x & y statements &
substitution or
elimination
Chapter review test
p.137,
#5 – 7
Chapter 1
Linear relations
Find the equation of
a line, given the
slope “M” and the y
intercept “B”
y = mx + b and algebra
Strategy
Read the problem carefully, consider the context and identify what stays
the same (constants) and what stays the same (variables)
Do let x represent and let y represent statements
One type of word problem lends itself well to the y = mx + b format – these
are usually the type where one variable depends on the other
Ex. Joe is comparing two internet service providers. Netaxes charges a flat
monthly fee of $10, plus $0.75 per hour spent on-line v. Webz charges a
flat
monthly fee of $5, plus $1 per hour. When do the plans cost the same?
Let x represent number of hours on-line and let y represent total cost
Equations would be y = 0.75x + 10 and y = x + 5
Other types of word problem lends itself well to the ax + by = c format
Ex. Premium gas sells for $78.9/L. Regular gas cells for $71.9/L. To boost
sales, middle octane gas is produced, and is sold at $73.9/L. If 1000 L of this
middle octane gas is produced, how much of each type of gas was used in the
mixture?
Let R represent the number of L of regular gas used
Let P represent the number of L of premium gas used
Equations would be R + P = 1000 and 71.9R + 78.9P = 739
(to set-up the 2nd equation you have to multiply 1000 times $73.9/litre)
Sometimes a chart helps you organize your information and helps set-up your
equations
Plug the slope in for m and your y-intercept in for b and you have your
equation, Ex. if slope is 3 and the y-intercept is 4, your equation is
y = 3x + 4
3
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 1
Linear relations
Chapter 1
Linear relations
If I want to… and
recommended ?’s
Find the equation of
a line, given the
slope “M” and a point
on the line that isn’t
the y intercept
Used in Chapter 2
verifying geometric
properties
Find the equation of
a line, given two
points
Used in Chapter 2
verifying geometric
properties
I will use…
Strategy
Y = mx + b and algebra
Substitute the slope and the point in for m, x and y and solve for b
Once you have b, rewrite the equation, putting the x and y back as
variables, not numbers
Y = mx + b and algebra
Calculate the slope using the slope formula from page 1
Substitute the slope and one of the points into y = mx + b and solve for b
Once you have b, rewrite the equation, putting the x and y back as
variables, not numbers
4
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 2
analytic
geometry
Chapter 2
analytic
geometry
If I want to… and
recommended ?’s
Find the radius of a
circle given the
centre at the origin
(0,0) and a point
Ch. review p.207,
#4
Find the equation of
a circle with the
centre at the origin,
given the radius
I will use…
Strategy
x2  y 2  r 2
Substitute the point in for x and y and solve for r
x2  y 2  r 2
Square r and plug in for r2
Ex. if r is 7 the equation of a circle would be
x 2  y 2  49
Ch. review p.207,
#5,6
Ch. review test
p.218, #2
Chapter 2
analytic
geometry
Chapter 2
analytic
geometry
Substitute the point coordinates in for x and y and solve for r 2
If r2 that you find is < the r2 in the given equation, then the point is inside
the circle
If r2 that you find is = the r2 in the given equation, then the point is on the
circle
If r2 that you find is > the r2 in the given equation, then the point is outside
the circle
Given an equation of
a circle with the
centre at the origin
(0,0), determine
whether a point is
inside, outside or on
the circle
p.156, #9
Find the midpoint of
a line, given two
points Ch. Review
p.211, #10-12 &
Ch. review test
p.218, p.1(a)
 x  x y  y2 
M 1 2, 1

2 
 2
Substitute the x and y values into the midpoint equation and simplify
5
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
If I want to… and
recommended ?’s
Chapter 2
analytic
geometry
Find the length of a
line segment
Chapter 2
analytic
geometry
Identify the
difference between
an altitude, median
and perpendicular
bisector of a
triangle
Chapter 2
analytic
geometry
Chapter 2
analytic
geometry
I will use…
Strategy
d  ( x2  x1 )2  ( y2  y1 )2
Substitute the x and y values into the length of a line segment equation and
simplify
Memorization
The altitude goes from the vertex of a triangle and meets the opposite
side at a 90° angle, but it doesn’t bisect the opposite side unless the vertex
is in an equilateral triangle or between the 2 equal sides of an isosceles
triangle
A median goes from the vertex of a triangle and bisects the opposite side
A perpendicular bisector bisects a side and goes straight up and either hits
another side, or in the case of a equilateral triangle, or an isosceles triangle
where the bisected side is the non-equal side, the vertex opposite
In an equilateral triangle, the altitude, median and perpendicular bisectors
are the same
y = mx + b
slope
finding equation of a
line
You have to find the equations of two median lines and then find the point
of intersection of those two lines
It can also be found by calculating the mean (average) of the x and y
coordinates of all three vertices
y = mx + b
slope
finding equation of a
line
You have to find the equations of the perpendicular bisectors of two sides,
then finding the point of intersection of those two lines
Ch. review p.209,
#7-9
See verify
geometric
properties for
application of this
Find the centroid of
a triangle
See verify
geometric
properties
Find the
circumcentre of a
triangle
See verify
geometric
properties
6
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 2
analytic
geometry
Chapter 2
analytic
geometry
If I want to… and
recommended ?’s
Find the
orthocenter of a
triangle
See verify
geometric
properties
Verify geometric
properties
Isosceles
triangle
Equilateral
triangle
Square
Rectangle
Rhombus
Parallelogram
Trapezoid
I will use…
Strategy
y = mx + b
slope
finding equation of a
line
You have to find the equations of two of the altitude lines and find the
point of intersection of those two lines
y = mx + b
slope
finding equation of a
line
Use properties of lines and line segments
Length of line segments can be verified or found using the distance formula
You can find whether line segments are parallel or perpendicular using the
slope formula and the negative reciprocal (flip and change the sign)
principal
Parallel lines have the same slope
Perpendicular lines have slopes that are the negative reciprocal of each
other
You may have to determine the coordinates of a point of intersection
before you can use the length of a line segment formula and the slope
formula to verify a geometric property
Length of a line
segment
d  ( x2  x1 )2  ( y2  y1 )2
Ch. review
p.213,#13-17 &
p.215,#18-19
Ch. review test
p.218,
#3,4,5,6,7,8
7
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
If I want to… and
recommended ?’s
I will use…
Expand and simplify
quadratic equations
Ch. review p.330,
#11 & Ch. review
test p.337,#5
Distributive law
Chapter 3 quadratic
equations
introduction
Identify whether a
relationship is linear
or quadratic
First and Second
Differences
Chapter 3 quadratic
equations
introduction
Find the common
factors of an
equation
Multiplication
Ch. review
p.332,#14,15
Basic Factoring Rules
Chapter 3 quadratic
equations
introduction
Or
FOIL
Strategy
With the distributive law, if working with a binomial, you multiply the first
term in the first bracket by each term in the second bracket. You then
multiply the second term in the first bracket by each term in the second
bracket
FOIL is similar and has an identical outcome, but goes FIRST, OUTSIDE,
INSIDE, LAST for the order of multiplying
Linear relations have straight lines, no x (exponents), rather just y and x,
and have first differences that are the same
Quadratic relations have curved lines in the shape of a parabola pointing up
2
or down, when not in factored form have a x (exponent), first differences
are different, but second differences are a constant or equal
2
Ch. review p.323,
#1-3, p.326, #46 & Ch. review test
p.337#1
When you are factoring, quadratics or otherwise, always see if there is a
common factor that you can factor out of the equation
Division
Ch. review test
p.337#6
8
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 3 quadratic
equations
introduction
Chapter 3 quadratic
equations
introduction
Chapter 3 quadratic
equations
introduction
If I want to… and
recommended ?’s
I will use…
Factor an equation in
Strategy
What two numbers add to give you B and multiply together to give you C
You may use the butterfly method also if you choose
form x  Bx  C
2
Ex. x  4 x  4 factors to (x + 2)(x + 2)
2
Ch. review p.
p.332,#14,15
Ch. review test
p.337,#6
Factor an equation in
form Ax  Bx  C
Ch. Rev.p.
p.332,#14,15 &
Ch. R.Tst p.337#6
Solve an equation in
2
Butterfly method
Or
Decomposition method
Butterfly method is the one that we did in class
See your notes
Algebra
Once you have found the factored form of the equation (which are set = to
0), set each bracket equal to 0 and solve for each bracket
Except in the case of the perfect square, ex. (x + 2)(x + 2) = 0, you should
get two values of x
Algebra
Once you have found the factored form of the equation (which are set = to
0), set each bracket equal to 0 and solve for each bracket
Except in the case of the perfect square, ex. (x + 2)(x + 2) = 0, you should
get two values of x
form x  Bx  C
(finding the zeros)
2
Ch. review
p.335,#16-20
Chapter 3 quadratic
equations
introduction
Ch. review test p.
Solve an equation in
form Ax  Bx  C
(finding the zeros)
2
Ch. review
p.335#16-20
Ch. review test
p.338,#7
9
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 3 quadratic
equations
introduction
Chapter 3 quadratic
equations
introduction
If I want to… and
recommended ?’s
I will use…
Strategy
Find factored form
quadratic equations
given roots and
optimal values
Factored form of quadratic equation is y  ( x  S )( x  T ) or
Ch. review
p.326,#4-6,
p.329,#8-10,
p.331,#12-13 &
Ch. review test
p.337,#3
can substitute the roots (for S and T), the optimal value of y and solve for
a
Once you have found the a, you rewrite the equation with the roots and the
a, but putting y back to the letter y, ex. y  2( x  3)( x  4)
Solving word
problems using
quadratic equations
or questions and
word problems, find
zeros, values of x
for diff. events and
optimal value &
graph/interpret
graphs & find
equations
Ch. review
p.326,#4-6,
p.329,#8-10,
p.331,#12-13 &
Ch. review test
p.338,#8-10 &
p.429#3,4
There are different variations of this, but basically, they use roots and
other properties to find sufficient information to find the equation
Ex. ball thrown in air question, zeros or x-intercepts, maximum height of
ball, when ball hits ground, time of maximum height
You usually find the roots first, the x values are the times they take place
Then, you can find the equation of the axis of symmetry by finding the
midpoint of the roots (take the average of the x values)
Then, you can plug the x value of the axis of symmetry into the equation
and get the optimal y-value, these two points together also give you the
coordinates of the vertex
y  a( x  S )( x  T )
Standard form of quadratic equation is
y  Ax 2  Bx  C
To find y  a( x  S )( x  T ) , if given the roots and the optimal value, you
10
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 4 –
vertex form
quadratics
If I want to… and
recommended ?’s
I will use…
Graph using
transformations
relative to
I will use the principles
of the vertex form of
the equation
y  x 2 (which has
y  a ( x  h) 2  k
vertex (0,0)
Ch. review p.418#
1-5, p.421,#6-8
Ch. review test p.
p.429#1,2,5
Chapter 4 –
vertex form
quadratics
Solve a word
problem where I am
asked to find the
maximum, minimum
or optimum value,
given information to
set-up the quadratic
in factored,
expanded form y=(xs)(x-t)
Ch. review
p.423,#10&11,
p.424#12-14
Ch. review test
p.429#6-7
Strategy
If h is –ve, the x coordinate of the vertex moves from 0 to the right by h
If h is +ve, the x coordinate of the vertex moves from 0 to the left by h
The y coordinate of the vertex moves from 0 up or down by k
If a is positive, the parabola opens up from the new vertex (h, k)
If a is negative, the parabola opens down from the new vertex (h, k)
If a is = 1, the graph is standard opening (move left and right from vertex
by one goes up/down by 1, move left and right from vertex by two goes
up/down by 4, etc.)
If a <1, the graph is a compressed or squished wider graph where the y
value is a x’s whatever the y value is on the standard opening graph (ex. if
a
I will first multiply the
factored form of the
equation using the
distributive law to get
it into standard form
1
, then move left and right from vertex by one goes up/down by 1/2,
2
move left and right from vertex by two goes up/down by 2, etc.)
If a >1, the graph is a stretched narrower graph where the y value is a x’s
whatever the y value is on the standard opening graph (ex. if a  2 , then
move left and right from vertex by one goes up/down by 2, move left and
right from vertex by two goes up/down by 8, etc.)
You could see this in the context of maximizing area, revenue, finding how
high a ball goes, etc.
Once you complete the square you can read off the vertex which provides
you with the maximum or minimum value and the x coordinate for that
maximum or minimum value
y  ax 2  bx  c
I will complete the
square to get the
equation into vertex
form
y  a( x  h)2  k from
which I can read the
vertex (h, k)
11
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 4 –
vertex form
quadratics
If I want to… and
recommended ?’s
Solve a word
problem where I am
asked to find what x
value will give me a
given y value given
infiormation to setup the quadratic in
factored or
expanded form y=(xs)(x-t)
Ch. review
p.427,#15-19
I will use…
I will first multiply the
factored form of the
equation using the
distributive law to get
it into standard form
y  ax 2  bx  c
Assuming that the
standard form isn’t
easily factorable I will
then use the quadratic
formula to find the
roots
x
Chapter 4 –
vertex form
quadratics
Find the roots of a
quadratic equation
given in vertex form
such as
Strategy
You’ll have to remember to move the y to the right hand of the equation
and subtract it from c
If the equation is easily factorable using the butterfly method or because
it is recognizable as a perfect square, then use those methods, rather than
the quadratic formula
(ex. find the dimensions that will give me an area of x, find the price of a
ticket that will generate a certain amount of revenue, find the time that
will give a certain height)
b  b2  4ac
2a
Set y = 0 and use
algebraic method to
solve for x
y  a( x  h)2  k or
y  ax 2  k
Chapter 4 –
vertex form
quadratics
Find how many roots
an equation has that
is in vertex form
y  a( x  h)2  k or
y  ax 2  k
Set y = 0 and us
algebraic method to
solve for x.
When you are solving using the algebraic method, you will have 1 of 3
possibilities
1. the number under the square root sign will be positive – 2 real roots
2. the number under the square root sign will be negative – no real roots
3. the number under the square root sign will be zero – 1 real root
12
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 4 –
vertex form
quadratics
If I want to… and
recommended ?’s
Find out how many
roots a quadratic
equation in the form
y  ax 2  bx  c has
I will use…
Use Discriminants
Calculate
b 2  4ac
Strategy
When you are solving using the algebraic method, you will have 1 of 3
possibilities
1. the number under the square root sign will be positive – 2 real roots
2. the number under the square root sign will be negative – no real roots
3. the number under the square root sign will be zero – 1 real root
13
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 5 –
primary trig
ratios
If I want to… and
recommended ?’s
Find out if 2
triangles are
congruent
I will use…
Strategy
The properties that all
corresponding angles
and sides are equal
All corresponding angles are equal
All corresponding sides are equal
The triangles may be rotated or flipped and still be congruent
The properties that all
corresponding sides
are proportional and
corresponding angles
are equal
All corresponding angles are equal
All corresponding sides are equal
The triangles may be rotated or flipped and still be congruent
Ch. review p.515,
#1
.
Chapter 5 –
primary trig
ratios
Chapter 5 –
primary trig
ratios
Find out if 2
triangles are similar
Ch. review p.p.515,
#2 & p.518#3-6 &
p.538,#1-3 &
p.586#1,2
Find missing sides or
angles in a right
angle triangle
SOHCAHTOA
AB BC AC


(ratio of corresponding sides)
DE EF DF
SinA 
OPP
ADJ
OPP
, CosA 
, TanA 
HYP
HYP
ADJ
Ch. review
p.522,#11-14
Ch. review test
p.526,#1abc,2-7
14
Grade 10 Academic Math – How Do I Chart – Mr. Baumgartner
Chapter
Chapter 6 – sine
law and cosine
law
If I want to… and
recommended ?’s
Find missing sides or
angles in a non-right
angle triangle
Ch. review
p.582,#4-6 &
p.585,#10-12
Chapter 6 – sine
law and cosine
law
Ch. review test
p.3-10
Find missing sides or
angles in a non-right
angle triangle
Ch. review
p.583,#7-9 &
p.585,#10-12
I will use…
Strategy
Sine law if given…
(1) 2 sides and one
angle across
from a known
side, or
(2) 2 angles and
any side
a
b
c


SinA SinB SinC
Cosine law if given…
(1) 2 sides and a
contained
angle, or
(2) 3 sides
a 2  b 2  c 2  2bc cos A,...or
b 2  a 2  c 2  2ac cos B,...or
c 2  a 2  b 2  2ab cos C ,...or
Any other combination of letters applied consistently
Ch. review test
p.3-10
15