Parabolas
Parabolas
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Parabolas
PARABOLAS
Quadratic equations have highest power 2 (squared). The standard form of a quadratic equation
is y = ax2 + bx + c
Answer these questions, before working through the chapter.
I used to think:
How many x-intercepts does a quadratic equation have?
What is the turning point of a parabola?
When is a turning point a maximum or minimum?
Answer these questions, after working through the chapter.
But now I think:
How many x-intercepts does a quadratic equation have?
What is the turning point of a parabola?
When is a turning point a maximum or minimum?
What do I know now that I didn’t know before?
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Parabolas
Basics
What is a Parabola?
A 'quadratic' equation has the form
y = ax2 + bx + c
This is called standard form
Where a, b and c are constants and a ! 0 . In other words, the highest power in a quadratic equation is 2 (squared).
Write these quadratic equations in standard form and find the values of a, b and c
a
y = 3x - x 2 + 4
y = x2 - 5
b
c
` y = -x 2 + 3x + 4
y = 3^ x - 5h^ x + 3h
` y = 3 (x2 - 2x - 15)
` y = 3x2 - 6x - 45
a
b
c
a
` a = -1, b = 3, c = 4
c
a
b
` a = 3, b = -6, c = -45
` a = 1, b = 0, c = -5
The graph of a quadratic equation is called a 'parabola'.
Drawing a Parabola from a Table
The table method has been used here to draw the graph of y = x2 - x - 2
x
-3
-2
-1
0
1
2
3
4
y
10
4
0
-2
-2
0
4
10
y
Each parabola has these key points:
10
9
• Two x-intercepts
8
• A y-intercept
7
6
• A turning point (or 'vertex')
5
4
3
2
x-intercept
x-intercept
1
0
-6 -5 -4 -3 -2 -1
-1
1
2
3
4
5
6
x -2
-3
y-intercept
2
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-4
Turning point
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c
Parabolas
Questions
Basics
1. Identify whether the following equations are quadratic or not:
a
y = 3x2 + 2x - 1
b
y = -2x + 1
c
y = x2 - 1
d
y = 1 + 2x + 4x2
e
y = 2x2 - 3x + 7
f
y = 3^ x - 2h^ x + 3h
2. Identify a, b and c in these quadratic equations:
a
y = x2 + 2x - 3
b
y = 3x2 - 5x + 4
c
y = 2 - 5x + 7 x 2
d
y = 2^ x - 1h^ x - 4h
e
y = -6x2 + 7
f
y = 2x^3x - 2h + 6
3. Find the x-intercepts and y-intercept of the following parabolas.
y
a
5
4
3
2
1
0 1 2 3 4
-4 -3 -2 -1
-1
-2
-3
-4
y
b
5
4
3
2
1
x 5
4
3
2
1
0 1 2 3 4
-4 -3 -2 -1
-1
-2
-3
-4
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c
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x 0 1 2 3 4
-4 -3 -2 -1
-1
-2
-3
-4
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x 3
Parabolas
Questions
Basics
4. Use the table below to draw the parabola of y = x 2 - 2 .
x
-3
-2
-1
0
1
2
3
y
y
6
5
4
3
2
1
-4
-3
-2
0
-1
1
2
3
4
x -1
-2
5. Use the table below to draw the parabola of y = x 2 .
x
-3
-2
-1
0
1
2
3
y
y
10
9
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7 8 9
-9 -8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
4
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Knowing More
It would be easier to draw parabolas if the key points could be found from the equation itself. These points are:
• The x-intercepts
• The y-intercept
• The turning point
• The leading coefficient a
Finding Intercepts
To find the x-intercepts, set y = 0 and solve the equation. To find the y-intercept, set x = 0 and solve for y.
Find the intercepts of the parabolas from the following quadratic equations:
a
y = x2 + 8x + 12
b
To find x-intercepts, set y = 0 and solve for x
y = 2x2 - 4x - 6
To find x-intercepts, set y = 0 and solve for x
2
x + 8x + 12 = 0
` (x + 6) (x + 2) = 0
2x2 - 4x - 6 = 0
` 2 (x + 1) (x - 3) = 0
` x = -6 or x = -2
` x = -1 or x = 3
` The x-intercepts are -6 and -2
` The x-intercepts are -1 and 3
To find y-intercepts, set x = 0 and solve for y
To find y-intercepts, set x = 0 and solve for y
y = 02 + 8 (0) + 12
` The y-intercepts is 12
= 12
y = 2 (02) - 4 (0) - 6
` The y-intercepts is -6
= -6
A parabola has only one y-intercept, but can have a maximum of two x-intercepts. This means it could have two,
one or even no x-intercepts. Here is a sketch of each case:
y
y
y
x Two x-intercepts:
The graph cuts the x-axis twice
x One x-intercept:
The graph touches the x-axis
x No x-intercept:
The graph does not touches the x-axis
The number of x-intercepts depends on how many solutions the quadratic equation has. If the equation is in
standard form y = ax2 + bx + c then the y-intercept is the value of c.
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Finding the Turning Point
The turning point of the parabola will have coordinates ^ x, yh . The x-coordinate can be found using
y = ax2 + bx + c and:
x =- b
This is the middle of the x-intercepts
2a
To find the y-value, substitute the above x-value into the equation. Here are some examples:
Find the turning points of these quadratic equations:
a
b
y = x2 + 4x - 21
a
b
y = 2x2 - 4x - 6
c
x-value: x =- b
2a
` x =- 4
2 (1)
` x = -2
a
c
b
x-value: x =2a
` x =- -4
2 (2)
`x=1
y-value: y = (-2) 2 + 4 (-2) - 21
y-value: y = 2 (1) 2 - 4 (1) - 6
` y = -25
b
` y = -8
` The turning point is (-2,-25)
` The turning point is (1,-8)
The turning point is also called the vertex. Here is an example how to draw a parabola from these key points:
Sketch the parabola for the quadratic equation y = 2x 2 - 4x - 6
Step 1: Find the intercepts and mark them off on the axes
y
x-intercept
2x2 - 4x - 6 = 0
` 2 (x + 1) (x - 3) = 0
` x = -1 or x = 3
x-intercept
y = 2 (02) - 4 (0) - 6
= -6
y-intercept
3
-1
x
y-intercept
-6
Step 2: Find the turning point, mark it on the graph and sketch the parabola
y
y = 2x2 - 4x - 6
x-value: x =- b
2a
` x =- - 4
2 (2)
`x=1
-1
y-value: y = 2 (1) 2 - 4 (1) - 6 = -8
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x
-6
(1,-8)
so the turning point is (1,-8)
6
3
Turning Point
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Parabolas
Knowing More
Effect of a
The value of a in y = ax2 + bx + c has an effect on what the parabola looks like. Up until now, all the examples
have had a positive value for a. A positive value for a causes the parabola to face upward and is called 'concave up'.
If a is negative then the parabola faces downwards and is called 'concave down'.
Concave up
Positive a (a 2 0)
Concave down
Negative a (a 1 0)
y
y
x
x
Whether the parabola is concave up ^a 2 0h or concave down ^a 1 0h is called the 'concavity' of the parabola
Basically, when drawing parabolas there are FOUR things to check:
1. The x-intercepts (set y = 0 , solve for x)
2. The y-intercept (set x = 0 , solve for y)
3. The turning point (x-value =- b , substitute into equation to find y-value)
2a
4. Whether it is concave up or concave down (is a 2 0 or a 1 0 ?)
a 2 0 , Happy parabola
a 1 0 , Sad parabola
Examples are on the next page.
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Sketch the graph of the following parabolas:
a
y = 2x2 + 4x - 16
b
1. Find the x-intercepts
y = -x2 - 4x + 5
1. Find the x-intercepts
2x2 + 4x - 16 = 0
- x2 - 4x + 5 = 0
` 2 (x - 2) (x + 4) = 0
` - (x - 1) (x + 5) = 0
` x-intercepts are 2 and -4
` x-intercepts are 1 and -5
2. Find the y-intercept
2. Find the y-intercept
y = - (0) 2 - 4 (0) + 5
2
y = 2 (0) + 4 (0) - 16
` y = -16
`y=5
` y-intercept is -16
` y-intercept is 5
3. Find the turning point
x-value: x =- b
2a
` x =- 4
2 (2)
` x =-1
3. Find the turning point
x-value: x =- b
2a
` x =- -4
2 (-1)
` x = -2
` y-value: y = - (-2) 2 -4 (-2) + 5
=9
` y-value: y = 2 (-1) 2 + 4 (-1) - 16
= -18
` the turning point is (-2, 9)
` the turning point is (-1,-18)
4. Check the Convavity
4. Check the Convavity
a = 2 ` a 2 0 (Concave up)
a = -1 ` a 1 0 (Concave down)
y
y
(-2, 9)
Turning Point
y-intercept
5
x-intercept
x
2
-4
-5
x
1
x-intercept
-16
(-1,-18)
Turning Point
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y-intercept
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Parabolas
Questions
Knowing More
1. Find the intercepts of the following quadratic equations by factorising or using the quadratic formula.
a
y = 2x2 - 6x - 8
b
x2 - 10x + 21
c
3x2 - 9x - 30
d
x2 + 6x - 24
2. Find the turning points of the above parabolas.
a
b
c
d
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Questions
3. Determine the number of x-intercepts and then find them (if possible).
a
y = x2 - 4x - 5
b
x2 - 4x - 21
c
x2 - 4x + 4
d
x2 + 5
e
x2 - 25
f
x 2 + 3x + 4
g
x2 + 8x + 16
h
-3x2 + 2x - 2
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Knowing More
Parabolas
Questions
Knowing More
4. Consider this quadratic equation y = 3x 2 - 6x - 9 :
a
Find the intercepts and mark them on the axes below.
b
Find the turning point and mark it on the axes below.
c
Draw the parabola.
y
12
11
10
9
8
7
6
5
4
3
2
1
-12-11-10-9 -8 -7 -6 -5 -4 -3 -2 -1-10
1 2 3 4 5 6 7 8 9 10 1112
x
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
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Questions
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5. Consider this quadratic equation y = 2x 2 - 8x + 8 :
a
How many x-intercepts does this parabola have? Mark off the x-intercepts and y-intercepts on the
axes below.
b
Find the turning point of this parabola. What do you notice about the turning point and the x-intercept?
c
Draw the parabola.
y
8
6
4
2
-4
-3
-2
0
-1
1
-2
-4
-6
-8
12
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3
4
x
Parabolas
Questions
Knowing More
6. What do we mean by 'concavity' of a parabola?
7. Identify if these parabolas are concave up or concave down.
a
y = x 2 - 5x - 6
b
y = -2x2 + 10x + 12
c
y = 36 - 3x2 - 3x
d
y = -5x + 2x2 - 12
e
y = 2 + x - 3x 2
f
y = (x + 7) (4x - 3)
8. Sketch the parabola of these quadratic equations – identify the intercepts and turning points on the axes.
a
y = 4x2 - 4x - 24
y
x
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b
Questions
Knowing More
y = -2x2 - 4x + 6
y
x
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Parabolas
Using Our Knowledge
The Axis of Symmetry
A vertical line exactly in the middle of a parabola's x-intercepts, is called the axis of symmetry or the axis of the
parabola. This line also passes through the turning point.
y
a20
Axis of Symmetry x = - b
2a
a10
Turning point
x
x
Turning point
Axis of Symmetry x = - b
2a
Since it is a vertical line, it has the equation x = k where k is a constant. There are two ways to find the value of k:
•
Method 1: The axis of symmetry passes through the turning point, so it's equation is
x =- b
2a
• Method 2: Since the axis of symmetry is exactly in the middle of the x-intercepts, the value of k can be
found from the average of the x-intercepts.
Find the axis of symmetry for the quadratic equation y = x 2 + 4x - 5
y
-5
x
1
x = -2
-5
Method 1
Method 2
Find the x-value of the turning point
x =- b
2a
` x =- 4
2 (1)
` x = -2
Find the x-intercepts:
` The axis of symmetry has equation x = -2
` The x-intercepts are at -5 and 1
The average of the x-intercepts is -5 + 1 = -2
2
` The axis of symmetry has equation x =- 2
x2 + 4x - 5 = 0
` (x + 5) (x - 1) = 0
x = -5 or x = 1
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Using Our Knowledge
Completing the Square
Completing the square can be used to find the axis of symmetry.
If a quadratic equation is in the form
y = a (x - p) 2 + q
then the turning point of the parabola is ^ p, qh .
Find the turning point of x 2 + 6x + 1
x2 + 6x + 1
Half
(x + 3) 2
Subtract
(x + 3 ) 2 - 9 + 1
Squared
(x + 3) 2 - 8
` x2 + 6x + 1 = (x + 3) 2 - 8
= (x - (-3)) 2 - 8
This must be
a minus "-"
p = -3
q = -8
` The turning point is (-3,-8)
Remember to find p, the negative value in the bracket is used.
Find the turning point and axis of symmetry of 3x 2 - 12x + 8
3x2 - 12x + 8
= 3 ` x2 - 4x + 8 j
3
= 3 8^ x - 2h2 - 4 + 8 B
3
= 3 8^ x - 2h2 - 4 B
3
= 3 ^ x - 2h2 - 4
p=2
q = -4
` The turning point is (2,-4)
` The axis of symmetry is x = 2
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Questions
Using Our Knowledge
1. Find the axis of symmetry of each of the following quadratic equations:
a
A equation with x-intercepts 4 and -2
b
A equation with x-intercepts -3 and 9
c
y = (x + 5) (x - 7)
d
y = -2 (x + 4) (x - 12)
e
y = x2 - 6x - 40
f
y = -2x2 + 7x + 15
g
y = 4x2 + 16x - 7
h
y = -3x2 + 18x - 14
2. Find the turning point of these parabolas:
a
y = (x - 3) 2 + 4
b
y = (x + 3) 2 + 4
c
y = 2 (x - 1) 2 - 3
d
y = -3 (x + 6) 2
e
y = -2x2 + 5 (hint: x2 = (x - 0) 2 )
f
y = (4 - x) 2 + 6
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Using Our Knowledge
3. Find the axis of symmetry of each of the following quadratic equations:
a
y = 2 (x - 4) 2 + 3
b
y = (x + 3) (x - 11)
c
y = x2 - 4x - 32
d
y = 2x2 - 16x + 33
e
y = -3 (x + 4) (x + 10)
f
y = -2x2 + 4x - 9
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Parabolas
Questions
Using Our Knowledge
4. Let’s say y = ^ x - Ah^ x - Bh .
a
What are the intercepts of this equation?
b
Write this quadratic formula in standard form.
c
Find the coordinates of the turning point in terms of A and B.
d
What is the axis of symmetry? Is this what you were expecting?
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Parabolas
Thinking More
The things to think about in this section are:
1. Does a parabola have a maximum or minimum point?
2. How do you find the equation from the parabola?
3. What happens when a parabola is shifted on the axes, horizontally or vertically?
Minimum/Maximum
In each parabola, the turning point is either the maximum or minimum. This depends on the concavity of the parabola.
Concave up
The turning point is the minimum
Concave down
The turning point is the maximum
y
y
Minimum
(Highest point)
x
x
Minimum
(Lowest point)
Answer these questions about y = 2x 2 + 4x + 3
a
Find the turning point of this parabola
x-coordinate =- b =- 4 = -1
2a
2 (2)
` y-coordinate = 2 (-1) 2 + 4 (-1) + 3 = 1
` The turning point is (-1, 1)
b
Is the turning point a maximum or minimum?
The equation of the parabola is y = 2x2 + 4x + 3
` a = 2 2 0 so this parabola is concave up.
` The turning point is a minimum
c
Use the turning point to show that this parabola will have no x-intercepts.
The turning point (-1,1) is a minimum, ` the parabola will not go below this point.
The y-coordinate (+1) is ABOVE the x-axis and since the parabola will not move below this point it will not
go through (or touch) the x-axis. ` there are no x-intercepts.
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1. When is the turning point a maximum and when is the turning point a minimum?
2. Find the turning point of the following parabolas, and state whether it is a maximum or minimum:
a
y = x2 + 6x + 5
b
y = -2x2 + 16x - 30
c
y = -x2 + 6x + 6
d
y = 3x2 + 30x + 73
3. Find the turning point of these parabolas and state whether or not they will have x-intercepts.
a
y = -x2 - 6x - 11
b
y = 2x2 + 20x + 51
c
y = 3x2 - 24x + 45
d
y = -2x2 - 16x - 26
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Finding the Equation from the Parabola
Up until now the focus has been how to draw the parabola from the functions (or equation). Finding the equation
from the parabola is also important. There are two main methods, and they are selected depending on what has
been given.
Method 1: You are given the turning point and another point
Let’s say the turning point ^ p, qh is given. Let's say there is another point ^ x, yh given on the parabola.
Step 1: Substitute the turning point ^ p, qh into the equation
y = a (x - p) 2 + q
Step 2: Substitute the point ^ m, nh for x and y and solve for a.
Find the equation of a parabola with turning point ^2, 9h which passes through the point ^1, 8h
p q
x y
Step 1: Substitute the turning point ^2, 9h into the equation
y = a (x - p) 2 + q
` y = a (x - 2) 2 + 9
Step 2: Substitute the point ^1, 8h for x and y and solve for a.
` 8 = a (1 - 2) 2 + 9
` a = 8-9
` a = -1
So, the equation of the parabola (in turning point form is):
y = - (x - 2) 2 + 9
y
(2, 9)
(1, 8)
x
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Method 2: You are given the x-intercepts and another point
Let's say the x-intercepts are x1 and x2 . Let’s say there is another point ^m, nh given on the parabola.
Step 1: Substitute the x-intercepts into the equation
y = a (x - x1) (x - x2)
Step 2: Substitute the point ^ m, nh for x and y and solve for a.
Find the equation of a parabola with x-intercepts at -2 and 1, if it also goes through the point ^-1,-4h
x1
m
y1
n
Step 1: Substitute the x-intercepts into the equation
y = a (x - x1) (x - x2)
` y = a (x - (-2)) (x - 1)
` y = a (x + 2) (x - 1)
` y = a (x2 + x - 2)
Step 2: Substitute in the given point, ^-1,-4h and solve for a.
` -4 = a ((-1) 2 + (-1) - 2)
` -4 = -2a
`a=2
So the equation of the parabola is:
y = 2 (x2 + x - 2)
` y = 2x2 + 2 - 4
y
-2
1
x
(-1,-4)
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4. Find the equations and draw a rough sketch of the following parabolas:
a
A parabola passing through ^-3, 4h with turning point ^-2, 3h .
b
A parabola with x-intercepts -5 and 4 which passes through the point ^3, 1h .
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Parabolas
Questions
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c
A parabola with turning point ^4,-2h and passing through the point ^6, 6h .
d
A parabola with x-intercepts -3 and 6 and passing through the point ^-1,-10h .
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5. Let’s say a parabola has the equation y = ax 2 + bx - 4 where a and b are unknown. However, it is known
that the parabola passes through the points ^-1, 1h and ^2, -2h .
a
Substitute the point ^-1, 1h into the parabola's equation to find a new equation in terms of a and b.
b
Substitute the point ^2,-2h into the parabola's equation to find a new equation in terms of a and b.
c
Use the equations from a and b above as simultaneous equations to solve for a and b.
d
Use a and b to write the complete equation of the parabola.
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6. Let’s say a parabola has the equation y = x 2 + bx + c where b and c are unknown. However, it is known that
the parabola passes through the points ^3, 16h and ^-2, 9h . Use the same process from the previous question
to find the complete equation of the parabola.
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Moving Parabolas Vertically
The graphs below are of y = x2 - 2 , y = x2 and y = x2 + 2 .
y
y = x2 + 2
y = x2
y = x2 - 2
2
x
0
-2
Adding units to a parabola shifts it up, and subtracting units from a parabola shifts it down.
Moving Parabolas Horizontally
The graphs below are of y = (x + 3) 2 - 1 , y = x2 - 1 and y = (x - 2) 2 - 1
y
2
y = (x + 3) - 1
y = x2 - 1
y = (x - 2) 2 - 1
-4
-3
-2
-1
1
2
3
4
x
-1
In any parabola, if x is replaced with x + a then the parabola is shifted a units to the left. In the above diagram:
• The parabola for y = (x + 3) 2 - 1 is 3 units to the left of y = x2 - 1 .
x has been replaced by x + 3
In any parabola, if x is replaced with x - a then the parabola is shifted a units to the right. In the above diagram:
• The parabola for y = (x - 2) 2 - 1 is 2 units to the right of y = x2 - 1 . This can be understood as -2 units to
the left.
x has been replaced by x - 2
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Here is an example of shifting parabolas:
The solid line below represents the quadratic equation y = (x - 1) 2 - 1 which can be simplified to y = x 2 - 2x
Find the equations of the dashed lines below:
y
10
9
8
7
6
f
5
a
4
3
2
e
1
-9 -8 -7 -6 -5 -4 -3 -2 -1
-1
c
1 2 3 4 5 6 7 8 9
-2
x
d
-3
-4
b
-5
-6
-7
-8
a
This graph is 5 units above y = (x - 1) 2 - 1 so, it must have the equation
y = (x - 1) 2 - 1 + 5 or y = x2 - 2x + 5
b
This graph is 4 units below y = (x - 1) 2 - 1 so, it must have the equation
y = (x - 1) 2 - 1 - 4 or y = x2 - 2x - 4
c
This graph is 5 units to the right of y = (x - 1) 2 - 1 so, it must have the equation
y = ((x - 5) - 1) 2 - 1 or y = x2 - 12x + 35
d
This graph is 5 units to the right and 2 units below y = (x - 1) 2 - 1 so, it must have the equation
y = ((x - 5) - 1) 2 - 1 - 2 or y = x2 - 12x + 33
e
This graph is 7 units to the left of y = (x - 1) 2 - 1 so, it must have the equation
y = ((x + 7) - 1) 2 - 1 or y = x2 + 12x + 35
f
This graph is 3 units to the left and 5 units above y = (x - 1) 2 - 1 so, it must have the equation
y = ((x + 3) - 1) 2 - 1 + 5 or y = x2 + 4x + 8
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7. The two parabolas below represent y = x 2 + 4x and y = -x 2 + 2x + 3 :
y
10
9
8
7
6
5
4
3
2
1
-9 -8 -7 -6 -5 -4 -3 -2 -1
-1
1 2 3 4 5 6 7 8 9
x
-2
-3
-4
-5
-6
-7
-8
a
Complete the square of each quadratic equation.
b
Use the above graphs to sketch the following on the same axes:
30
y = x2 + 4x - 3
y = -x2 + 2x + 6 (Hint: y = -x2 + 2x + 3 + 3 )
y = - (x - 1) 2 + 2 (Hint: y = - (x - 1) 2 + 4 - 2 )
y = (x + 3) 2 + 4 (x + 3)
y = - (x - 2) 2 + 2 (x - 2) + 3
y = - ((x + 5) - 1) 2 + 6
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Basics:
Basics:
1. a Yes
b
No
c
Yes
Yes
e
Yes
f
Yes
d
2. a
a = 1, b = 2, c =-3
b
a = 3, b =-5, c = 4
c
a = 7, b =-5, c = 2
d
a = 2, b =-10, c = 8
e
a =-6, b = 0, c = 7
f
a = 6, b =-4, c = 6
5.
3. a The x-intercepts are at: x =-2 and x = 4
The y-intercept is at: y =-4
4.
b
The x-intercepts are at: x =-1 and x = 2
The y-intercept is at: y =-4
c
The x-intercepts are at: x = 0 and x = 3
The y-intercept is at: y = 0
x
-3
-2
-1
0
1
2
3
y
7
2
-1
-2
-1
2
7
x
-3
-2
-1
0
1
2
3
y
9
4
1
0
1
4
9
Knowing More:
1. a The x-intercepts are 4 and -1
The y-intercept is -8
b
The x-intercepts are 3 and 7
The y-intercept is 21
c
The x-intercepts are 5 and -2
The y-intercept is -30
d
The x-intercepts are x =-3 + 33 or
x =-3 - 33
The y-intercept is -24
2. a The turning point is ` 3 , -12 1 j
2
2
b
c
d
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The turning point is ^5, -4h
The turning point is ` 3 , -36 3 j
2
4
The turning point is ^-3, -33h
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Knowing More:
Knowing More:
5.
3. a There are two x-intercepts at 5 and -1
b
There are two x-intercepts at -3 and 7
c
There is only one x-intercept at 2
d
There is no x-intercept
e
There are two x-intercepts at 5 and -5
f
There is no x-intercept
g
There is only one x-intercept at -4
h
There is no x-intercept
6. The concavity of a parabola is the direction the
parabola faces, either up or down. In the form of
y = ax2 + bx + c , if a is positive ^a 2 0h , then
the concave faces up, if a is negative ^a 1 0h
then the concave faces down.
4. a The x-intercept are x = 3 and x =-1
The y-intercept is -9
b
c
The turning point is ^1, -12h .
c
7. a a is positive, ` concave up
b
a is negative, ` concave down
c
a is negative, ` concave down
d
a is positive, ` concave up
e
a is negative, ` concave down
f
a is positive, ` concave up
8. a
5. a There is one x-intercept at 2
The y-intercept is 8
b
The turning point is ^2, 0h .
The turning point is at the x-intercept.
-2
3
-24
1
` 2 , -25 j
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Knowing More:
Using Our Knowledge:
3. a The axis of symmetry has equation x = 4
8. b
(-1, 8)
y = 6
x = -3
b
The axis of symmetry has equation x = 4
c
The axis of symmetry has equation x = 2
d
The axis of symmetry has equation x = 4
e
The axis of symmetry has equation x =-7
f
The axis of symmetry has equation x = 1
x = 1
4. a The x-intercepts of the function are
x = A and x = B
Using Our Knowledge:
1. a The axis of symmetry has equation x = 1
b
b
The axis of symmetry has equation x = 3
c
c
The axis of symmetry has equation x = 1
d
d
The axis of symmetry has equation x = 4
e
The axis of symmetry has equation x = 3
f
The axis of symmetry has equation x = 7
4
g
The axis of symmetry has equation x =-2
h
2. a
b
c
d
e
f
y = x2 - ^ B + Ah x + AB
c B + A,c
2
-^ B + Ah2
+ AB mm
4
The axis of symmetry is the x-value at the
turning point, x = B + A
2
This is to be expected as it is the same as
the average of the two x-intercepts, A and B
Thinking More:
1. The turning point will be a maximum when
the parabola is concave down (ie when a is
negative).
The axis of symmetry has equation x = 3
The turning point is ^3, 4h
The turning point will be a minimum when the
the parabola is concave up, when a is positive.
The turning point is ^-3, 4h
The turning point is ^1, -3h
2. a Because the value of a = 1 2 0 (ie a is
positive), the parabola is concave up and
the turning point` a minimum at ^-3, -4h
The turning point is ^-6, 0h
The turning point is ^0, 5h
b
Because the value of a =-2 1 0 (ie a is
negative), the parabola is concave down and
the turning point ` a maximum at ^4, 2h
c
Because the value of a =-1 1 0 (ie a is
negative), the parabola is concave down and
the turning point ` a maximum at ^3, 15h
The turning point is ^4, 6h
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Thinking More:
Thinking More:
2. d Because the value of a = 3 2 0 (ie a is
positive), the parabola is concave up and
the turning point ` a minimum at ^-5, -2h
4.
c
y = 2x2 - 16x + 30
30
3. a The turning point is ^-3, -2h
The parabola will not have x-intercept
b
The turning point is ^-5, 1h
The parabola will not have x-intercept
c
The turning point is ^4, -3h
The parabola will intercept the x-axis
d
The turning point is ^-4, 6h
The parabola will intercept the x-axis
4. a
^6, 6h
^4, -2h
d
y = 5 ^ x2 - 3x - 18h
7
y = x2 + 4x + 7
7
-3
6
^-1, -10h
-12 6
7
5. a a = b + 5
a
y = - 1 ^ x2 + x - 20h
8
b
2a + b = 1
c
a=2
b =-3
5
2
-5
d
^3, 1h
y = 2x2 - 3x - 4
4
2
6. y = x + 2 x + 5 4
5
5
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Thinking More:
7. a
b
Thinking More:
y = ^ x + 2h2 - 4 and y =-^ x - 1h2 + 4
7. b
y = - (x - 1) 2 + 2
y = x2 + 4x - 3
y = (x + 3) 2 + 4 (x + 3)
y = -x2 + 2x + 6
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Thinking More:
7. b
y = - (x - 2) 2 + 2 (x - 2) + 3
y = - ((x + 5) - 1) 2 + 6
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