Download MPM 2D1 - Mr. Tjeerdsma

MFM 2P1
Name: __________
Graphing Quadratics Using x & y Intercepts
1. Make a sketch of the following parabola
o y-intercept of 8
o x-intercepts of -4 and -2
o -4 and -2 are also called the ____________
The standard form of the equation for
this parabola is y = x2+6x+8
The factored form of the equation for
this parabola is y=(x+4)(x+2)
2. Looking at the sketch for question 1, we notice the value of the ____ variable is zero for the
x-intercepts.
3. Looking at the sketch for question 1, we notice the value for the ____ variable is zero for the
y-intercept.
Part A: We can find the y-intercept of the parabola from the standard form very quickly!
Example
Your Turn!
Find the y-intercept of the parabola
with the equation y=x2+4x-12.
Find the y-intercept of the parabola with
the equation y=x2-8x+12.
y-intercept  x=0
y-intercept  x= _____
y = x2 + 4x - 12
y = (0)2 + 4(0) – 12
y = -12
y= x2 - 8x + 12
Therefore the y-intercept is (0,-12)
y= ____
y=( ___ )2 -8 ( ___ ) + 12
Therefore the y-intercept is ( ___ , ___ )
Without doing any calculation determine the y-intercept of the following parabola’s by looking at
their equation in standard form (y=ax2+bx+c).
y = 3x2+3x+9
y-int = ( ___, ___)
y = x2–3x–12
y-int = ( ___, ___)
y = x2+12x+11
y-int = ( ___, ___)
Part B: We can find the x-intercept of the parabola from the factored form very quickly!
When does the product (multiplication) of two numbers equal ZERO?
5 x ___ = 0
___ x 7 = 0
-3 x ___ = 0
In order for the product of two numbers to equal zero one of the number must equal _______.
We can use this to find the x-intercepts of a parabola when it’s equation is in factored form
Example
Your Turn!
Find the x-intercept of the parabola
with the equation y=(x-6)(x+2)
Find the x-intercept of the parabola
with the equation y=(x-6)(x-2)
x-intercept  y=0
x-intercept  y= ____
y = (x-6)(x+2)
0 = (x-6)(x+2)
This product
equals zero
Therefore;
(x-6)=0 or (x+2) = 0
x-6 = 0
x+2 = 0
x=6
x = -2
y = (x-6)(x-2)
___ = (x-6)(x+2)
Therefore;
______ = 0 or _____ = 0
As a result are x-intercepts are
(6,0) and (-2,0)
=
=
=
=
As a result are x-intercepts are
(___ , 0) and (___, ___)
Using what you have learnt so far during the investigation complete the following table without
graphing
Standard Form
Factored Form
Example:
y = x2+7x+10
y = (x+2)(x+5)
x-intercepts (zeros)
y-intercept
x + 2 = 0 or x + 5 = 0
y = 02 +7(0) + 10
x = -2
x = -5
x-ints: (-2,0) & (-5,0)
y = x2 + 4x - 21
y = (x-3)(x+7)
y = x2 – x – 21
y = (x+4)(x-5)
y = 10
y-int: (0,10)
10
Part 3: Application
y
9
8
Use what you have learned to sketch the following
parabolas:
Parabola #
Standard Form
7
6
5
4
Factored Form
3
2
1
1.
y=
x2
+ 2x – 8
y = (x + 4)(x - 2)
x
0
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1 -1 0
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
-2
-3
2.
y=
x2
– 6x + 5
-4
y = (x – 5)(x – 1)
-5
-6
-7
3.
y=
x2
-9
-8
y = (x + 3)(x - 3)
-9
-10
10
Parabola #
Standard Form
Factored Form
y
9
8
7
6
4.
5
y = (x - 8)(x + 1)
4
3
2
1
x
0
-10
-9
-8
-7
-6
-5
-4
-3
-2
y = (x + 3)(x – 2)
5.
-1 -1 0
-2
-3
-4
-5
-6
-7
6.
-8
y = (x - 3 )(x - 2)
-9
-10
10
Parabola #
Standard Form
Factored Form
y
9
8
7
6
7.
5
y = x2 - 2x - 3
4
3
2
1
x
0
-10
8.
-9
-8
-7
-6
-5
-4
-3
-2
-1 -1 0
-2
y=
x2
-4
-3
-4
-5
-6
-7
9.
y=
x2
- 4x + 4
-8
-9
-10