Section C
Non Linear Graphs
Graphic Calculators will be
useful for this topic
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Some words to learn
Plot a graph: Draw graph by plotting points
Sketch/Draw a graph: Do not plot, draw free hand but make
sure the graph goes through certain points
Intercept: Where a graph crosses an axis
Linear graph: Straight line graph (y = mx + c)
Non linear graph: Curved graph
Quadratic: Expression with highest power of squared
Parabola: U shape graph
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Some more words to learn
Graph transformation: Moving a graph or changing its shape
Vertex of graph: Top/Bottom of parabola
Origin: Where the axis of a graph meet
Factorised: Put into brackets
Critical points: points on a graph of interest i.e. where it
crosses the axes
Translation: Sliding a graph up/down/left/right
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Non-linear graphs
Examples of non-linear functions include,
y = 2x + x8
y = x2 + 1
y = 7x3 – 3x
y=
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5
x–2
–6
y = 3 + 2x
SLO
Plotting graphs
Graphing equations by plotting points
http://www.youtube.com/watch?v=mYYGQrRQeUw
5 of 48
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Plotting graphs using a table of values
Plot the graph of
y = x2 – 3
for values of x between –3 and 3.
We can use a table of values to generate coordinates that lie
on the graph as follows:
x
–3
–2
–1
0
1
2
3
y = x2 – 3
6
1
–2
–3
–2
1
6
(–3, 6) (–2, 1) (–1, –2)(0, –3) (1, –2) (2, 1) (3, 6)
Take care with graphic calculators and negative numbers squared
6 of 48
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x
–3
–2
–1
0
1
2
3
y = x2 – 3
6
1
–2
–3
–2
1
6
The points given in the
table are plotted …
… and the points are then
joined together with a
smooth curve.
The shape of this graph is
called a parabola.
It is characteristic of a
quadratic.
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y
5
4
3
2
1
–3
–2
–1
0
–1
–2
1
2
3
x
Plotting graphs using a table of values: Example 2
y = x3 – 7x + 2
Plot the graph of
for values of x between –3 and 3.
This function is more complex and so it is helpful to include
more rows in the table to show each stage in the substitution.
x
–3
–2
–1
0
1
2
3
x3
–27
–8
–1
0
1
8
27
– 7x
+ 21
+14
+7
+0
–7
– 14
– 21
+2
+2
+2
+2
+2
+2
+2
+2
y = x3 – 7x + 2
–4
8
8
2
–4
–4
8
8 of 48
Plotting graphs using a table of values
x
–3
–2
–1
0
1
2
3
y = x3 – 7x + 2
–4
8
8
2
–4
–4
8
y
Plot the points given in the
table.
10
8
Joined points together with
a smooth curve. (Merit
students: you might need to
plot extra points at vertices.
6
4
2
–3
The shape of this graph is
characteristic of a cubic.
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–2
–1
0
–2
–4
1
2
3
x
Your Turn
Plot the graph y = x2 (Use x values from -3 to 3)
x
y = x2
-3
9
-2
4
-1
1
0
0
1
1
2
4
3
9
y
10
9
8
7
6
5
4
3
2
1
– 6– 5– 4– 3– 2–– 11
– 2
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TIP FOR PLOTTING
We can see that the shape is
up 1 out 1
up 3 out 1
up 5 out 1
up 7 out 1 etc.
1 2 3 4 5 6 x
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Your Turn
Plot the graph y = –x2 (Use x values from -3 to 3)
x
y = -(x2)
-3
-2
-1
0
1
2
3
-9
-4
-1
0
-1
-4
-9
y
2
1
– 6– 5– 4– 3– 2–– 11
– 2
– 3
– 4
– 5
– 6
– 7
– 8
– 9
– 10
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1 2 3 4 5 6 x
The graph is the
same shape as x2
but upside-down
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Your Turn
Plot the graph y = x2 + 5 (Use x values from -3 to 3)
x
–3
–2
–1
0
1
2
3
y = x2 + 5
14
9
6
5
6
9
14
y
15
12
9
6
3
–3
–2
–1
0
–3
–6
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1
2
3
x
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Your Turn
Plot the graph y = x3 + 2x + 10 from -3 to 3
x
–3
–2
–1
0
1
2
3
y = x3 + 2x
-23
-2
7
10
13
22
43
y
50
40
30
20
10
–3
–2
–1 0
–10
–20
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1
2
3
x
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Your Turn
Plot the graph y = 2x + 6 from -3 to 3
–3
x
y = x3 + 2x
–2
6.125 6.25
y
–1
0
1
2
3
6.5
7
8
10
14
This is called an exponential
Graph. To learn more go to the
following links
15
12
9
6
3
–3
–2
–1
0
–3
–6
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1
2
3
http://www.youtube.com/watch?v=9SOSfRNCQZQ
http://www.youtube.com/watch?v=0uMBIDRv3bI
x
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Questions to do from the books
Achieve
Merit
Excellence
Gamma
EAS
15 of 48
P37 Q138 – 141
P39 Q152 – 158
P71 Q211 – 218
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Know basic non linear graph shapes
16 of 48
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Quadratic graphs
A quadratic always contains a term in x2 as its highest
exponent. It can also contain terms in x or a constant.
y = x2
y = x2 – 3x
y = –3x2
The shape of a quadratic graph is called a parabola
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Cubic graphs
A cubic always contains a term in x3 as its highest exponent.
y = x3 – 4x
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y = x3 + 2x2
y = -3x2 – x3
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Exponential graphs
An exponential equation is always of the form y = ax,
where a is a positive number.
y = 2x
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y = 3x
y = 0.25x
Exploring exponential graphs
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Your Turn:
Give a possible equation of the following graphs
y = x3 + …
21 of 48
y = 2x + …
y = x2 + …
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Sketching Parabolas
y = x2
Tip for sketching:
To sketch a parabola, find
the vertex then use the
following.
Out 1
Out 1
Out 1
Out 1
etc.
up
up
up
up
y
7
11
10
9
8
7
1
3
5
7
5
6
5
4
3
3
2
This graph has vertex at (0 , 0)
1
-3
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-2
-1
1
–6
1
2
3
x
SLO
Translating a graph up and down
Translating graphs up and down
http://www.youtube.com/watch?v=5K5GG5Az288 (+C)
http://www.youtube.com/watch?v=dKIaPwJeX48&list=UU_kZjlWQbAW45Ly8PdQ3Hw&index=34&feature=plcp (-C)
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Vertical translations
Here is the graph of y = x2.
This is the graph of y = x2 + 1
y
and this is the graph of y = x2 + 4.
What do you notice?
x
This is the graph of y = x2 – 3
and this is the graph of y = x2 – 7.
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What do you notice?
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Translating graphs up and down
y=
y
2
x
±a
+
–
5
E.g. Adding a number to x2 slides the
graph up.
4
3
2
1
–3
–2
–1
0
–1
–2
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1
2
3
x
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E.g. The graph of y = x2 – 2 has the same shape as x2 but
with the vertex at y = –2 (click to show new graph)
y = x2 – 2
y
y = x2
y
–3
–2
–1
5
5
4
4
3
3
2
2
1
1
0
–1
–2
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1
2
3
x
–3
–2
–1
0
–1
–2
1
2
3
x
Set coefficient of x as 0 and then change constant to see
effect of graph
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Your Turn
Sketch the following graphs. Remember sketch does not
mean a plot.
–3
y = x2 + 1
y
y = x2 – 3
y
5
5
4
4
3
3
2
2
1
1
–2
–1
0
–1
–2
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1
2
3
x
–3
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–2
–1
0
–1
–2
1
2
3
x
Set coefficient of x and x2 as 0 and then change constant to see effect of graph
29 of 48
Questions to do from the books
Achieve
Gamma
P116 Ex9.01 Q1 – 4
EAS
P45 Q159, 162
30 of 48
Merit
Excellence
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Translating a graph left and right
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Horizontal translations
Here is the graph of y = x2 .
This is the graph of y = (x – 1)2
y
and this is the graph of y = (x – 4)2
What do you notice?
x
This is the graph of y = (x + 2)2
and this is the graph of y = (x + 3)2
What do you notice?
32 of 48
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Translating graphs left and right
+
y = (x ± a )2
–
E.g. Adding a number to x2 inside the brackets
slides the graph left e.g. (x + 5)2 slides the
parabola 5 to the left.
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E.g. The graph of y = (x + 3)2 has the same shape as x2 but
with the vertex at x = -3 (click to see other graph)
y = x2
y
–3
–2
–1
5
5
4
4
3
3
2
2
1
1
0
–1
–2
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y = (x + 3)2
y
1
2
3
x
–3
–2
–1
0
–1
–2
1
2
3
x
Your Turn
Sketch the following graphs
y = (x – 1)2
y
–3
–2
–1
5
5
4
4
3
3
2
2
1
1
0
–1
–2
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y = (x + 2)2
y
1
2
3
x
–3
–2
–1
0
–1
–2
1
2
3
x
Questions to do from the books
Achieve
Gamma
P116 Ex9.01 Q1 – 4
EAS
P45 Q160, 161
36 of 48
Merit
Excellence
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Reflecting a graph in the x axis
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Reflecting a graph in the x axis
y = –x2
Placing a negative sign in front of an equation
reflects it in the x axis (flips it up-side-down)
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y = x2
y
E.g. The graph of y = –x2 is an upside down U.
(click to see new graph)
y = –x2
y
5
5
4
4
3
3
2
2
1
1
–3
–2
–1
0
–1
1
2
3
–3
x
–2
–1
0
–1
–2
–2
–3
–4
39 of 48
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–5
–6
1
2
3
x
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Combining transformations
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We have learnt 3 transformations
Up/down
y = x2 + c ……………. slides up by c
y = x2 – c ……………. slides down by c
Left/right
y = (x + c)2 ……………. slides left by c
y = (x – c)2 ……………. slides right by c
Reflection
y = –x2 .. Reflects in the x axis (upside down U shape)
41 of 48
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Combining Transformations
Transformation can be combined.
e.g. y = –(x + 2)2 + 3 has three transformations
The negative
turns the U
upside down
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The +2 inside
the bracket
slides it 2 left
The +3 at the
end slides it up
by 3
Combining Transformations
E.g. Sketch y = (x – 3)2 – 4
y
5
4
Start with parabola: U
3
2
1
Slide 3 right
–3
–2
–1
0
–1
–2
Slide down by 4
–3
–4
43 of 48
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–5
–6
1
2
3
x
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Combining Transformations
E.g. Sketch y = –(x + 2)2 + 3
y
5
4
Start with an upside
down U
3
2
1
Slide 2 left
Slide up by 3
–3
–2
–1
0
–1
–2
–3
–4
44 of 48
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–5
–6
1
2
3
x
Your Turn
Describe the transformation of each parabola.
Equation
Transformations
7 DOWN
y = x2 – 7
y = (x + 5)2 + 6 5 LEFT and 6 UP
y = –x2 – 4
UPSIDE DOWN and 4 DOWN
UPSIDE DOWN and 4 RIGHT
y = –(x – 4)2
y = (x + 2)2 – 8 2 LEFT and 8 DOWN
y = 8 – x2
UPSIDE DOWN and 8 UP
y = 8 + (x – 7)2 7 RIGHT and 8 UP
45 of 48
Your Turn
Sketch the following graphs
y = (x – 2)2 + 3
y
–3
46 of 48
–2
–1
y = –(x + 1)2
y
5
5
4
4
3
3
2
2
1
1
0
–1
1
2
3
x
–3
–2
–1
0
–1
–2
–2
–3
–3
–4
–4
–5
–5
–6
–6
1
2
3
x
Your Turn
Sketch the following graphs
y = –x2 + 1
y = (x + 1)2 – 2
y
y
–3
47 of 48
–2
–1
5
5
4
4
3
3
2
2
1
1
0
–1
1
2
3
x
–3
–2
–1
0
–1
–2
–2
–3
–3
–4
–4
–5
–5
–6
–6
1
2
3
x
Questions to do from the books
Achieve
Merit
Excellence
P117 Ex9.01 Q13 – 15
Gamma P117 Ex9.01 Q5 – 12 P119 Ex9.02 Q4, 5
P124 Ex9.06
EAS
P46 Q163 – 166
Merit students: do a couple of
achieve questions from above
and then do the next few slides
48 of 48
Important practice
for exams at Merit
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Finding equations of the form y = ±(x ± a)2 ± b
given the parabola
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Finding equations of the form
y = ±(x ± a)2 ± b given the parabola
Merit students are expected to
work backwards. E.g. find
equation of parabola on right.
y
5
4
3
This x2 graph has three transformations
2
1
Upside down: negative
–3
2 left: (x + 2)
0
–1
–3
Equation of graph is……
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–1
–2
5 high: + 5
y = –(x +
–2
2)2
+5
–4
–5
–6
1
2
3
x
Your Turn: Find the equation of the following graphs
–3
51 of 48
y = x2 – 3
y = (x – 2)2 – 4
y
y
–2
–1
5
5
4
4
3
3
2
2
1
1
0
–1
1
2
3
x
–3
–2
–1
0
–1
–2
–2
–3
–3
–4
–4
–5
–5
–6
–6
1
2
3
x
Your Turn: Find the equation of the following graphs
y = –(x – 2)2 + 3
y = –(x + 1)2
y
y
–3
52 of 48
–2
–1
5
5
4
4
3
3
2
2
1
1
0
–1
1
2
3
x
–3
–2
–1
0
–1
–2
–2
–3
–3
–4
–4
–5
–5
–6
–6
1
2
3
x
Questions to do from the books
Achieve
Merit
Excellence
P117 Ex9.01 Q13 – 15
Gamma P117 Ex9.01 Q5 – 12 P119 Ex9.02 Q4, 5
P124 Ex9.06
EAS
P46 Q163 – 166
Important practice
for exams at Merit
53 of 48
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Finding intercepts of a graph in the form
y = ±(x ± a)(x ± b) or y = ±x(x ± a)
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x and y intercepts
The following graph has three Intercepts (places
where it crosses an axis).
y
5
4
3
2
1
–3
–2
–1
0
–1
–2
55 of 48
1
2
3
x
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Finding intercepts
To find the x intercept substitute y = 0 into equation
To find the y intercept substitute x = 0 into equation
E.g. find the intercepts of the graph y = (x + 5)(x – 3)
To find y intercept substitute x = 0
y = (0 + 5)(0 – 3) = (5) x (–3) = –15
Crosses the y
axis at –15
To find x intercept substitute y = 0
0 = (x + 5)(x – 3)
The trick here is to change the sign of the number inside
each bracket i.e. –5 and 3
Crosses the x
axis at –5 and 3
56 of 48
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Now we can sketch the graph ….
y
6
3
Crosses the x
axis at –5 and 3
–6
–4
–2
0
2
4
6
-3
-6
-9
–12
–15
Crosses the y
axis at –15
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x
Finding intercepts: Example 2
E.g. find the intercepts of the graph y = (x + 2)(x – 7)
To find y intercept substitute x = 0
y = (0 + 2)(0 – 7) = (2) x (–7) = –14
Crosses the y
axis at –14
To find x intercept substitute y = 0
0 = (x + 2)(x – 7)
The trick here is to change the sign of the number inside
each bracket i.e. –2 and 7
Crosses the x
axis at –2 and 7
58 of 48
Now we can sketch the graph ….
y
6
3
Crosses the x
axis at –2 and 7
–6
–4
–2
0
2
4
6
-3
-6
-9
–12
–15
Crosses the y
axis at –14
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x
Finding intercepts: Example 3
E.g. find the intercepts of the graph y = –(x + 3)(x – 6)
(notice this is an upside down U)
To find y intercept substitute x = 0
y = –(0 + 3)(0 – 6) = –(3) x (–6) = 18
Crosses the y
axis at 18
To find x intercept substitute y = 0
0 = –(x + 3)(x – 6)
The trick here is to change the sign of the number inside
each bracket i.e. –3 and 6
Crosses the x
axis at –3 and 6
60 of 48
Now we can sketch the graph ….
y
18
15
12
9
6
Crosses the x
axis at –3 and 6
3
–6
–4 –2
0
x
2
4
–3
Crosses the y
axis at 18
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6
Finding intercepts: Example 4
E.g. Find the intercepts of the graph y = x(x + 8)
To find y intercept substitute x = 0
y = 0(0 + 8) = (0) x (–7) = 0
Crosses the y
axis at 0
To find x intercept substitute y = 0
0 = x(x + 8)
The trick here is to change the sign of the number inside
the bracket i.e. –8 and remember that an x on its own
means it will cross at 0
Crosses the x
axis at 0 and –8
62 of 48
Your Turn
Find the intercepts of the following parabolas
Equation
y = (x + 4)(x – 2)
y = (x + 9)(x – 7)
y = (x + 4)(x + 5)
y = –(x – 5)(x – 1)
y = – (x – 7)(x + 3)
y = x(x + 6)
y = x(x + 2)
y = –x(x + 8)
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y intercept
x intercept
–8
–4 and 2
–63
–9 and 7
20
–4 and –5
–5
5 and 1
21
7 and –3
0
0 and –6
0
0
0 and –2
0 and –8
Your Turn: sketch the following graphs
Equation
y = (x + 4)(x – 2)
y = (x + 4)(x + 5)
y = –(x – 5)(x – 1)
64 of 48
y intercept
x intercept
–8
–4 and 2
20
–4 and –5
–5
5 and 1
Your Turn: Answer 1
Equation
y = (x + 4)(x – 2)
y intercept
x intercept
–8
–4 and 2
y
20
15
10
5
–6
–4
–2
0
–5
–10
65 of 48
2
4
6
x
Your Turn: Answer 2
Equation
y = (x + 4)(x + 5)
y intercept
x intercept
20
–4 and –5
y
20
15
10
5
–6
–4
–2
0
–5
–10
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2
4
6
x
Your Turn: sketch the following graphs
Equation
y = –(x – 5)(x – 1)
y intercept
x intercept
–5
5 and 1
y
20
15
10
5
–6
–4
–2
0
–5
–10
67 of 48
2
4
6
x
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Finding the vertex of a graph
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Vertex of a graph
The vertex of a graph is the bottom of the U or
the top if the U if up side down
y
–3
–2
–1
y
5
5
4
4
3
3
2
2
1
1
0
–1
–2
69 of 48
1
2
3
x
–3
–2
–1
0
–1
–2
1
2
3
x
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y
The x coordinate of
the vertex for a
parabola is always
half way between the
two x intercepts
5
4
3
2
1
E.g. in this graph the
vertex is found half
way between –3 and 1
i.e. x = –1
–3
Click to show this graphically
70 of 48
–2
–1
0
–1
–2
1
2
3
x
Your Turn
Find the x coordinate of the vertex of the following quadratics
Equation
x coordinate of vertex
–1
y = (x + 4)(x – 2)
y = (x + 9)(x – 7)
–1
–4.5
y = (x + 4)(x + 5)
y = –(x – 5)(x – 1)
3
y = –(x – 7)(x + 3)
2.5
Click for
–3
y
=
x(x
+
6)
graphic
calculator y = x(x + 2)
–1
–4
y = –x(x + 8)
71 of 48
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Y coordinate of mid point
To find the y coordinate of the vertex, substitute the
coordinate of the x midpoint into the equation
E.g. The x mid point of y = (x + 4)(x – 8) is 2 (half way
between –4 and 8)
Substitute –2 into y = (x + 4)(x – 8)
y = (–2 + 4)(–2 – 8)
y = (2) x (–10)
y = –20
Vertex is at (–2, –20)
72 of 48
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Y coordinate of mid point: Example 2
E.g. The x mid point of y = (x – 6)(x – 10) is 8 (half way
between 6 and 10)
Substitute 8 into y = (x – 6)(x – 10)
y = (8 – 6)(8 – 10)
y = (2) x (–2)
y = –4
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Vertex is at (8, –4)
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Find the y coordinate of the vertex of the following quadratics
Your Turn
Equation
y = (x + 4)(x – 2)
y = (x + 9)(x – 7)
y = (x + 4)(x + 5)
y = (x – 5)(x – 1)
y = (x – 7)(x + 3)
y = x(x + 6)
y = x(x + 2)
y = –x(x + 8)
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x coordinate
–1
–1
–4.5
3
2.5
–3
–1
–4
y coordinate
–9
–64
–0.25
–1
–1
–1
–1
–1
SLO
Sketch graphs of the form y = ±(x ± a)(x ± b) or
y = ±x(x ± a) using intercepts and vertex
75 of 48
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Sketching graphs using intercepts and vertex
To sketch a graph, plot the intercepts
and vertex and freehand draw the rest.
y
15
E.g.
Sketch the graph y = (x + 3)(x – 5)
12
9
Step 1: Find the intercepts
Crosses y axis at: –15
Crosses x axis at –3 and 5
Step 2: Find the vertex
x coordinate of vertex is: 1
y coordinate of vertex is: –16
6
3
–6
–4
–2
0
–3
–6
–9
Step 3: Plot the points
–12
Step 4: freehand sketch the graph
–15
–18
76 of 48
2
4
6
x
Your Turn:
Sketch the graph of the following quadratic y = (x + 4)(x – 2)
y
Step 1: Find the intercepts
Crosses y axis at: –8
Crosses x axis at –4 and 2
15
12
9
6
Step 2: Find the vertex
x coordinate of vertex is: –1
y coordinate of vertex is: –9
3
–6
–4
–2
0
–3
–6
Step 3: Plot the points
–9
–12
Step 4: freehand sketch the graph
–15
–18
77 of 48
2
4
6
x
Your Turn:
Sketch the graph of the following quadratic y = –x(x + 4)
y
Step 1: Find the intercepts
Crosses y axis at: 0
Crosses x axis at 0 and –4
15
12
9
6
Step 2: Find the vertex
x coordinate of vertex is: –2
y coordinate of vertex is: 4
3
–6
–4
–2
0
–3
–6
Step 3: Plot the points
Step 4: freehand sketch the graph
78 of 48
Notice negative makes an
upside down U
–9
–12
–15
–18
2
4
6
x
Your Turn:
Sketch the graph of the following quadratic y = –(x + 6)(x – 2)
y
Step 1: Find the intercepts
Crosses y axis at: 12
Crosses x axis at –6 and 2
15
12
9
6
Step 2: Find the vertex
x coordinate of vertex is: –2
y coordinate of vertex is: 16
3
–6
–4
–2
0
–3
–6
Step 3: Plot the points
–9
–12
Step 4: freehand sketch the graph
–15
–18
79 of 48
2
4
6
x
Questions to do from the books
Achieve
Merit
Gamma
P122 Ex 9.04
Q1 – 5, 7 – 9
P123 Ex 9.04 Q10
P124 Ex 9.06
EAS
P49 Q167 – 174
80 of 48
Excellence
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graphic
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SLO
Stretching graphs
(Sketching and finding equations)
Stretching y =x2
http://www.youtube.com/watch?v=tjroyVI8El4&feature=related
81 of 48
Graph stretches
y
This is the graph of y = x2
This is the graph of y = 2x2
This is the graph of y = 3x2
What do you notice?
x
82 of 48
The bigger the number in front of x2
the more the graph is stretched. Or in
more detail, the number in front of x2
is the amount it is stretched up by
y   x2
What do you notice?
The bigger the number in front of x2
the more the graph is stretched.
y  2 x 2
y  3x 2
y  5 x 2
y yy
2
1
– 6
83 of 48
– 5
– 4
– 3
– 2
– 1
– 1
– 2
– 3
– 4
– 5
– 6
– 7
– 8
– 9
– 10
– 11
– 12
– 13
– 14
– 15
1
2
3
4
5
6
x xx
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y  x2
1 2
x
2
1
y  x2
4
1
y  x2
10
y
y   x2
What do you notice?
1
y   x2
2
1
y   x2
4
1
y   x2
yyyy 10
10
9
8
7
6
5
4
3
2
1
– 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 ––11
– 2
– 3
– 4
– 5
– 6
– 7
– 8
– 9
– 10
84 of 48
1
2
3
4
5
6
7
8
9
10
xxxx
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Set constant and coefficient of x as 0. Use blue arrow to see how the
change in the coefficient of x2 changes the graph
85 of 48
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Stretching graphs parallel to the y axis
y
A number in front of an
equation (i.e. y = kx2)
stretches the graph parallel
to the y axis by a factor of k
x
86 of 48
SLO
Finding equation with stretch factor given parabola
The following is Merit work
Find equation given parabola
http://www.youtube.com/watch?v=DrHgyPFR_0s
87 of 48
Find equation: Stretch example
Find the equation of the parabola
To make it easier draw
the graph of y = x2 so we
can make a comparison
10
8
Step 1: Identify the basic
equation i.e. y = x2.
6
Step 2: Place k in front
of equation to allow for
the stretch. i.e. y = kx2
2
Step 3: Choose any
point from the graph i.e.
(2, 8)
4
–4
–3
–2
–1 0
–2
–4
–6
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1
2
3
4
Step 4: Substitute
values into y = kx2 and
solve to find k
10
8
6
8 = k22
8 = 4k
Therefore k = 2
4
2
Step 5: Substitute
values of k into y = kx2
–4
–3
–2
–1 0
–2
–4
y = 2x2
89 of 48
–6
1
2
3
4
Copy into
your notes
Stretch example 2
Find the equation of the parabola
10
Step 1: Identify the basic
equation i.e. y = x2 – 6
8
6
Step 2: Place k in front
of equation to allow for
the stretch.
i.e. y = kx2 – 6
Step 3: Choose any
point from the graph i.e.
(2, 4)
4
2
–4
–3
–2
–1 0
–2
–4
–6
90 of 48
1
2
3
4
Copy into
your notes
Step 4: Substitute
values into y = kx2 – 6
and solve to find k i.e.
10
8
6
4 = k22 – 6
4 = 4k – 6
Therefore k = 2.5
4
2
Step 5: Substitute value
of k into y = kx2 – 6
–4
–3
–2
–1 0
–2
–4
y = 2.5x2 – 6
91 of 48
–6
1
2
3
4
Your Turn
Find the equation of the parabola
10
Step 1: Identify the basic
equation
i.e. y = –(x – 2)2 + 8
8
6
Step 2: Place k in front
of equation to allow for
the stretch.
i.e. y = –k(x – 2)2 + 8
Step 3: Choose any
point from the graph i.e.
(4, 6)
4
2
–4
–3
–2
–1 0
–2
–4
–6
92 of 48
1
2
3
4
Step 4: Substitute
values into
y = –k(x – 2)2 + 8
and solve to find k
6 = –k(4 – 2)2 + 8
6 = –k22 + 8
6 = –4k + 8
–2 = –4k
Therefore k = 0.5
10
8
6
4
2
–4
Step 5: Substitute value
of k into y = –k(x – 2)2 + 8
y = 0.5(x – 2)2 + 8
93 of 48
–3
–2
–1 0
–2
–4
–6
1
2
3
4
Your Turn
Find the equation of the parabola
10
Step 1: Identify the basic
equation (not clear
vertex!) so use
i.e. y = (x + 1)(x – 3)
Step 2: Place k in front
of equation to allow for
the stretch.
i.e. y = k(x + 1)(x – 3)
Step 3: Choose any
point from the graph i.e.
(2, –4)
8
6
4
2
–4
–3
–2
–1 0
–2
–4
–6
94 of 48
1
2
3
4
Step 4: Substitute
values into
–4 = k(2 + 1)(2 – 3)
and solve to find k
–4 = k(3)(–1)
–4 = –3k
4
Therefore k = 3
Step 5: Substitute value of
k into y = k(x + 1)(x – 3)
10
8
6
4
2
–4
–3
–2
–1 0
–2
–4
y=
95 of 48
4
(x
3
+ 1)(x – 3)
–6
1
2
3
4
Questions to do from the books
Achieve Merit
Gamma
P119 Ex9.02 Q1,2,6
P124 Ex 9.06
P122 Ex 9.04 Q5, 9, 11, 12
EAS
P39 Q152 – 158
P53/54 Q175 – 172
96 of 48
Excellence
P60 Q189–196
(merit/excellence)
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SLO
Sketching un-factorised quadratics of
the form ±x2 + bx + c
Merit/excellence work
This involves higher algebra skills than most students have at the
moment (we have not done the algebra topic yet). We will do the
notes but leave the questions for now.
97 of 48
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your notes
Sketching un-factorised quadratics
To sketch a parabola the quadratic needs to be factorised first.
E.g. Sketch y = x2 + 5x + 6
Step 1: Factorise
y = x2 + 5x + 6 = (x + 2)(x + 3)
Step 2: Sketch as usual by finding intercepts and vertex
x intercept = –2 and –3
y intercept = 6
Vertex = (2.5, 24.75)
98 of 48
Questions to do from the books
Achieve Merit
Gamma
P122 Ex 9.04 Q6
EAS
P56 Q183 – 188
99 of 48
Excellence
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Solving equations using graphs
Merit work
100 of 48
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your notes
Using graphs to solve equations
To solve the equation below consider the left-hand side and
the right-hand side of the equation as two separate equations.
10
3x + 1 = 6x – 2
8
y = 3x + 1
6
y = 6x – 2
4
Draw these graphs.
2
The x-coordinate(s) of
where these two graphs
intersect will give the
solutions to the equation.
–4
–3
–2
–1 0
–2
–4
x=1
101 of 48
–6
1
2
3
4
Using graphs to solve equations: Example 2
10
y = 2x2 – 5
y = 3x
8
(2.5, 7.5)
6
–4
–3
Solve 2x2 – 5 = 3x
The graphs of
y = 2x2 – 5 and y = 3x
intersect at the points:
4
(–1, –3)
2
and (2.5, 7.5).
–2
–1 0
–2
(–1,–3)
–4
–6
1
2
3
4
The x-value of these
coordinates give us the
solution to the equation
2x2 – 5 = 3x as
x = –1
and x = 2.5
102 of 48
Using graphs to solve equations: Example 3
Solve the equation x3 – 3x = 1 using graphs.
This equation does not have any exact solutions and so the
graph can only be used to find approximate solutions.
A cubic equation can have up to three solutions and so the
graph can also tell us how many solutions there are.
Again, we can consider the left-hand side and the right-hand
side of the equation as two separate functions and find the xcoordinates of their points of intersection.
x3 – 3x = 1
y = x3 – 3x
103 of 48
y=1
Using graphs to solve equations
10
y = x3 – 3x
The graphs of y = x3 – 3x
and y = 1 intersect at
three points:
8
This means that the
equation x3 – 3x = 1 has
three solutions.
6
4
2
–4
–3
–2
–1 0
–2
–4
–6
104 of 48
y=1
1
2
3
4
Using the graph these
solutions are
approximately:
x = –1.5
x = –0.3
x = 1.9
Questions to do from the books
Achieve Merit
Gamma
Excellence
P123 Ex 9.04 Q14 – 16
EAS
105 of 48
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SLO
Applications of quadratics
Merit/excellence work
Application problems
http://www.youtube.com/watch?v=vAPPYoBV2Ow
106 of 48
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Applications of quadratics
Convert the words into facts and then draw a rough sketch.
Assume all projectiles follow a parabolic path (e.g. the kick
of a ball follows an upside down U shape)
The vertex of all parabolas is the maximum/minimum
107 of 48
Applications of quadratics: Example 1
The length of the base and the height of this rectangle add to 11
Find the greatest possible area.
x
11-x
1
10
9
8
2
3
5
7
6
6
5
30
7
9
4
3
2
28
24
18
10
1
10
8
11  x
10
18
24
28
30
4
y
Area
x
x  y  11
y  11  x
A  xy
A  x(11  x)
108
108 of 48
35
5
10
15
20
25
30
1
2
3
4
5
6
7
8
9
10
11
A
10
18
24
28
30
30
28
24
18
10
We need to try the point between 5 and 6
Greatest possible area
A  x(11  x)
A  (5.5)(11  5.5)
y  30.25
35
Area of Paddock m2
x
1
2
3
4
5
6
7
8
9
10
30
25
20
15
10
5
1
109 of 48
2
3
4
5
6
7
8
Side length m
9
10
11
x
A farmer is planning to construct a rectangular paddock
Example 3
using 10m of fence. One side is along a river and doesn't
need a fence. What is the maximum area of the paddock?
x  2.5
y  10  2(2.5)
y 5
A  2.5  5
 12.5
centre
13
12
11
10
9
8
7
6
1
5
4
3
2
1
2
3
4
5
6
7
8
9
10
11
12
13
x  x  y  10 54321
2 x  y  10
y  10  2 x
A  xy
A  x(10  2 x)
x
x
y
y
13
12
x
y
A
11
1
8
8
9
Greatest Area
10
12.5m 2
8
2
6
12
3
4
12
6
4
2
8
4
5
0
0
7
5
3
2
1
110 of 48
1
2
3
4
5
x
110
Applications of quadratics: Example 4
Bob kicks a ball. It reaches a height of 25m and travels 70m.
Give the equation for the height of the ball.
h
25m
70m d
The basic equation of the ball is h = –k(d – 35)2 + 25
When the ball has gone 70m, the height is 0.
i.e. 0 = -k(70 – 35)2 + 25, Solve and therefore k = 0.02…
Equation is
h = –0.02(d – 35)2 + 25
(Where h is height and d is horizontal distance)
111 of 48
Method 2: Applications of quadratics: Example 4
Bob kicks a ball. It reaches a height of 25m and travels 70m.
Give the equation for the height of the ball.
25m
70m
The basic equation of the ball is y = –kx(x – 70)
The maximum height of the ball is half way between 0 and 70
i.e. 35 (this gives us values to find k)
25 = –k35(35 – 70) therefore k = 0.02
Equation is
h = –0.02d(d – 70)
(Where h is height and d is horizontal distance)
112 of 48
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Applications of quadratics: Example 5
A mixing bowl is in the shape of a parabola. The bowl has a
diameter of 30cm and a depth of 25cm. Give the equation for the
depth of the bowl.
30cm
– 25cm
The basic equation of the bowl is y = kx(x – 30)
The maximum depth of the bowl is half way between 0 and 30
i.e. 15 (this gives us values to find k)
–25 = k15(15 – 30) therefore k = 0.02
Equation is
D = 0.11d(d – 30)
(Where D is depth and d is distance along diameter)
113 of 48
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Another way ……
A mixing bowl is in the shape of a parabola. The bowl has a
diameter of 30cm and a depth of 25cm. Give the equation for the
depth of the bowl.
25cm
30cm
The basic equation of the bowl is y = k(x – 15)2
Use right hand side to find point we know i.e. (30, 25)
Substitute these values in to find k
25 = k(30 – 15)2 therefore
k = 0.02
Equation is
D = 0.11(d – 15)2
(Where D is depth and d is distance along diameter)
114 of 48
Excellence ….
(Probably too hard until algebra skills improve)
There are now 2 equations for the bowl
D = 0.11d(d – 30)
D = 0.11(d – 15)2
Show that both these bowls have exactly the same shape
D = 0.11d(d – 30) = 0.11d2 – 3.33d
D = 0.11(d – 15)2 = 0.11d2 – 3.33d + 25
The second of the two equations is exactly the same apart from
it has been translated 25 units up i.e. it is the same shape
115 of 48
Scale factors and finding equations of Parabolas
987654321– 10
10
11
12
13
14
15
987654321
To Find the Equation of a Parabola
87654321– 21
We can find the equation from 1) the vertex and another point
2) x intercepts and another point
Eg1)
We can see that the vertex is (3,  10)
So equation must be
y
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
x y
sub in the point (1,0)
y  k ( x  3) 2  10
0  k (1  3)2  10
A clear point is (1,0)
0  k (2)2  10
0  k (4)  10
10
 10
So equation is
y  2.5( x  3) 2  10
116 of 48
10  4k
10
k
4
k  2.5
–2
–1–1
–2
–3
–4
–5
–6
–7
–8
–9
– 10
1
2
3
4
5
6
7
8
x
2. We could also find the equation using the x intercepts
(1,0)(5,0)
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
 y  k ( x  1)( x  5)
x
y
We may then use the vertex as a point (3, 10)
So sub in the point
y  k ( x  1)( x  5)
10  k (3  1)(3  5)
10  k (2)(2)
10  4k
10
4
k  2.5
–2
so equation is
y  2.5( x  1)( x  5)
k
y
–1–1
–2
–3
–4
–5
–6
–7
–8
–9
– 10
1
Note: y  2.5( x  3) 2  10 is equal to
y  2.5( x  1)( x  5)
(Will show this soon)
117 of 48
2
3
4
5
6
7
8
x
So why have we got 2 different equations for the same parabola?
y  2.5( x  1)( x  5)
y  2.5( x  3)2  10
y  2.5( x  1)( x  5)
y  2.5( x  3) 2  10
y  2.5( x 2  5 x  x  5)
y  2.5( x  3)( x  3)  10
y  2.5( x  6 x  5)
y  2.5( x 2  3x  3x  9)  10
2
y  2.5 x  15 x  12.5
2
y  2.5( x 2  6 x  9)  10
y  2.5 x 2  15 x  22.5  10
y  2.5 x 2  15 x  12.5
We can see that these are both the same equation
118
118 of 48
Either of the following 3 ways is acceptable
y
8
7
6
5
4
3
2
1
x y
1)Vertex is (  3,8) a clear point is (1,0)
y  k ( x  3)2  8
y  0.5( x  3)2  8
0  k (1  3)  8
8
8
2
8  k (4)2
8  16k
k  0.5
2) Again we could have used the x intercepts
x
y
(1,0) and (-7,0) and the point (-3,8)
– 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1– 1
–2
–3
–4
–5
–6
–7
–8
–9
– 10
-7
0
8  k (3  1)(3  7)
-3
8
1
0
8  16k
k
119 of 48
y  0.5( x  1)( x  7)
8
 k  0.5
16
2
3
3) On calc
y  k ( x  1)( x  7)
8  k (4)(4)
1
y  0.5 x 2  3x  3.5
We can see either way the scale factor
is the same
4
x
Questions to do from the books
Achieve
Gamma
EAS
120 of 48
Merit
Excellence
P125 Ex 9.07
P66 Q204 – 210
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graphic
calculator
The Exam
The examination papers from previous years
appear to concentrate on 3 main aspects:
1) Find equation of parabola using y = ±(x ± a)2 + b
2) Knowing that the midpoint between x intercepts
is where to find the lowest/highest point.
3) Sketching parabolas rather than plotting.
121 of 48
Random Revision questions
122 of 48
Find the equations of the following graphs
yy
y  ( x  3)2  4
10
9
8
7
y  ( x  2)2  5
6
5
4
y  ( x  1)2  2
3
2
1
– 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1
– 1
– 2
– 3
– 4
– 5
123 of 48
1
2
3
4
5
6
x
y  ( x  4)2  1
Find the equations of the following graphs
yy
5
4
y   x2  4
3
2
1
–6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6
7
8
xx
y  ( x  5)2
–2
–3
–4
–5
–6
–7
–8
–9
– 10
124 of 48
y  ( x  2)2  3
Sketch
3)
Plot
y
10
y  x 3
2
y  x
8
6
4
2
y  ( x  4)
2
2
–4
–2
2
–2
–4
–6
–8
– 10
125 of 48
4
6
8
x
Sketch y = (x – 3)(x – 5)
y
10
8
6
4
2
–2
2
4
6
8
10
x
–2
–4
126
126 of 48
y
10
Sketch
1) Plot
8
3 x  4 y  20  0
6
y  x2  4x  5
4
y   ( x  4) 2
2
y  ( x  2) 2  5
y  x 1
y  ( x  1)( x  3)
2
–8
–6
–4
–2
2
–2
–4
–6
–8
– 10
127 of 48
4
6
8
y
1Sketch
) Plot
10
2 y  5 x  20
8
y  x2  8x  7
6
y  ( x  5) 2
4
y  ( x  1) 2  4
2
y   x2  2
y  ( x  2)( x  4)– 8
–6
–4
–2
2
–2
–4
–6
–8
– 10
128 of 48
4
6
8
x
y
1)
Graph
Sketch
3
y   x 5
4
y  ( x  5) 2  1
14
y  ( x  1)( x  3)
4
12
10
8
6
2
– 10
–8
–6
–4
–2
–2
–4
–6
–8
– 10
– 12
– 14
129 of 48
2
4
6
x
(Merit question)
1) Find the equation of the graph
y
14
y  k ( x  4)( x  2)
4  k (4  4)(4  2)
12
10
4  k (8)(2)
8
4  16k
6
k  0.25
 y  0.25( x  4)( x  2)
4
2
– 10
–8
–6
–4
–2
2
–2
–4
–6
130 of 48
point (4,4)
4
6
8
10
x
y
10
1)
Plot
Sketch
8
y  x2  6
6
y  ( x  1) 2
4
y   ( x  4) 2
2
y  ( x  2) 2  5
–8
–6
–4
–2
2
y   x2  1
–2
y  ( x  1)( x  3)
–4
–6
–8
– 10
131 of 48
4
6
8
y
10
Sketch
1) Plot
2 y  5 x  20
8
y  x2  8x  7
6
y  ( x  5) 2
4
y  ( x  1) 2  4
2
y   x2  2
–8
y  ( x  2)( x  4)
–6
–4
–2
2
–2
–4
–6
–8
– 10
132 of 48
4
6
8
Sketch
8)
Graph
y  ( x  3)(1  x)
y
4
2
– 4
– 2
2
– 2
– 4
– 6
– 8
– 10
– 12
– 14
133 of 48
4
6
8
10
x