Last Thursday, we were discovering the relationship between the equation and the graph.
1. By looking at the equation of the quadratic, how can you determine the direction of opening of the graph?
2. By looking at the equation of the quadratic, how can you determine the zero(s)?
x‐intercept(s)
3. How can the zeros help you determine the coordinates of the vertex?
4. How many distinct zeros can a quadratic relation have?
Recall: the zeroes of a parabola occur at points on the x axis (ie where the y coordinate equals 0).
Apr 5­7:50 PM
Formats for a Quadratic Equation
• factored form: y=a(x­s)(x­t)
In the factored form, • a tells us • s and t are To find the vertex, we
The optimal value is y‐coordinate of the vertex
(maximum or minimum).
• standard form: y=ax2 + bx + c
• vertex form: y=a(x­h)2+k Apr 5­8:55 PM
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Solving for zeros algebraically when in factored form:
y = a(x‐s) (x‐t)
Since the zeros occur at the x‐axis, set the y‐value
of an equation to 0 and solve for the corresponding x's.
Notice that when the product of two numbers equals 0, at least one of the numbers must equal 0. 0 = a(x‐s)(x‐t); a cannot be zero by definition
That means that either x‐s = 0 or x‐t = 0
x = s or
x=t
Therefore, s and t are the zeros.
Apr 5­8:51 PM
Example
• Determine the zero(s) for each of the following and use them to find the coordinates of the vertex.
• Write each equation in factored form.
1. y=x(x­10)
2. y=x(30­x)
Factored
form
Another method to determine the zeros:
Vertex
Apr 5­9:08 PM
2
Example
• Determine the zero(s) for each of the following and use them to find the coordinates of the vertex.
• Write each equation in factored form.
3. y=­(x­2)(x­8)
4. y=(3+x)(2­x)
Find the zeros
Method 1:
Notice that this is already in factored
form. Then the zeros are 2 and 8.
Method 2:
Set y=0 since we are looking for the x‐
intercepts.
0 = ‐(x‐2) (x‐8)
0 = (x‐2) (x‐8) multiplied both sides by ‐1
Either x‐2 = 0 or x‐8 = 0
x = 2 or
x=8
To get into factored form:
write with x's at
y = (x+3)(‐x+2)front of bracket
factor out ‐1 from 2nd
y = (x+3)(‐1)(x‐2) bracket to get +x
y = ‐1(x+3)(x‐2) move ‐1 to front. It is
actually the coefficient a
Therefore, zeros are 2 and 8.
To find the vertex,
h = (2+8)/2
find the midpoint of
the zeros (x‐value)
h=5
k = ‐(5‐2)(5‐8)
=9
y = ‐(x+3)(x‐2)
find the optimal
value (y‐value)
Therefore the vertex is (5,9).
Apr 5­9:09 PM
Example
• Determine the zero(s) for each of the following and use them to find the coordinates of the vertex.
• Write each equation in factored form.
5. y=­2(x­5)(3x­1)
6. y=2(x­2)2
To determine zeros:
let y=0 for x‐intercepts
0 = ‐2(x‐5)(3x‐1)
0 = (x‐5)(3x‐1) divide both
zeros: 2,2
There is only 1 distinct zero (2)
sides by ‐2
To determine vertex:
The midpoint between the
zeros is just 2.
vertex (2, 0)
Either x‐5=0 or 3x‐1=0
x=5 or x=1/3
zeros: 5, 1/3
Vertex (8/3, ‐97/3)
To factored form:
y = ‐2(x‐5)(3x‐1)
factor out the 3 from the
y = ‐2(x‐5) (3)(x‐1/3) 2nd bracket to get +x at
the front
y = ‐6(x‐5)(x‐1/3)
Factored form:
y = 2(x‐2)2
or
y = 2(x‐2)(x‐2)
Apr 5­9:09 PM
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Example
• Determine the zero(s) for each of the following and use them to find the coordinates of the vertex.
• Write each equation in factored form.
7. y=5x2 ­15x
8. A=24L­L2
To determine zeros:
let y=0 for x‐intercepts
0 = 5x2 ‐ 15x
0 = 5x (x‐3) need to factor
0 = x (x‐3)
Either x=0 or x‐3=0
x=3
zeros: 0, 3
Vertex (1.5, ‐11.25)
To factored form:
y = 5x2 ‐ 15x
y = 5x(x ‐3)
To determine zeros:
let A=0 for L‐intercepts
0 = 24L‐L2
0 = L(24‐L)
Either L=0 or 24‐L=0
L=24
zeros: 0, 24
Vertex (12, 144)
To factored form:
A = 24L ‐ L2
A = L(24‐L)
A = L(‐L+24)
A = ‐L(L‐24)
Apr 5­9:10 PM
HW: p. 267 # 11, 14, 15
p. 230 # 5ace, 6adgjnq
Apr 5­9:11 PM
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