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Unraveling the
Practice Test
Shift 4 to right
and up 2
Check each answer:
22  5(2)  14 4-10-14 =-20
7 2  5(7)  14 49-35-14 =0
(2) 2  5(2)  14 4+10-14 =0
(7) 2  5(7)  14 49+35-14 =70
x
x
x
x
When
Convert
multiplying
from Radical
with
To exponential
the same base
form
add3 all
1 exponents:
𝑥 =the
𝑥 3
1
Fractions
𝑥 =need
𝑥 2
common1denominators:
1
− 2
1 =𝑥
𝑥 2
𝑥
1
3
•𝑥
3
2
•
51
−
−
𝑥 𝑥 62
=
1
(1 3+3 2−5 6− )
2
𝑥
=
3
(2 6+9 6−5 6−6)
𝑥
=
3
11 −8 )
(
6
6
𝑥
=𝑥 6
=𝑥
1
2
1
• 𝑥−
2
1
2
(x+1)
(x-1)
(x-2)
(x-2)
(x+2)
(x+3)
(x-3)
x=-1
(x+4)
(x-4)
x=2
Odd number of factors
Fits with graph’s end behavior
x=-3
x=4
So (x+3) is a factor
So (x+1) is a factor
So (x-2) is a (repeated) factor {since it’s a turning point}
So (x-4) is a factor
𝒙≥𝟐
𝒙≥𝟐
𝒙+𝟑≥ 𝟐+𝟑
(𝒙 + 𝟑) ≥ 𝟓
−𝟐 𝒙 + 𝟑 ≥
≤ −𝟐(𝟓)
Switch the inequality
When you multiply
Or divide by a negative
≤
Rate of change = slope
25 x +
Charge for zero days
y-intercept
2 5 x + 30
= Total
2 5 x + 30 = 1 8 0
f(x)
g(x)
Graphing the functions will be difficult
and doesn’t show the area of the graph
we want.
f(x)
g(x)
Graphing the functions will be difficult
and doesn’t show the area of the graph
we want.
f(x)
<
<
<
<
<
<
>
>
>
g(x)
Graphing the functions will be difficult
and doesn’t show the area of the graph
we want.
f(x) < g(x) from x=0 to 5 and then
f(x) > g(x) from x=6 to 8
So the two functions must intersect
somewhere between 5 and 6
1 00 0
1000
x
-
50
1000
x
-
50
=
1 00 0
x+ 1
Type equation here.
64 = 8 = x-3
+3
+3
=> X = 11
20 = 4 × 5 = 2 5
1
𝑥=
2
16 4 1
= =
8 2
8
+3
+3
39 = 𝑥 + 3
Type equation here. 3𝑥 2 = 752
Type equation here. 2𝑥 = 10
2
𝑥− 5=2 5
1
𝑥 =
2
2
39 = (𝑥 + 3)
2
2
𝒙=
𝒙=𝟑 𝟓
𝟏
𝟒
39 = 𝑥 2 + 6𝑥 + 9
0 = 𝑥 2 + 6𝑥 − 30
3𝑥 = 5625
𝑥 = 1875
𝑥=5
Doesn’t factor, so not integer
Opens UP
(-5)
(-4)
Vertex =(-3,-2)
(-3)
“Roots” are also called
Zeros or solutions.
Real zeros are x-intercepts
(-2)
(-1)
0
1
2
3
4
5
Two Roots
Opens DOWN
(-5)
(-4)
Vertex =(-2,3)
(-3)
“Roots” are also called
Zeros or solutions.
Real zeros are x-intercepts
(-2)
(-1)
0
1
2
3
4
5
Two Roots
Opens DOWN
(-5)
(-4)
Vertex =(-2,-3)
Sandy’s claim
(-3)
(-2)
(-1)
0
1
2
3
4
5
No Roots
Opens UP
(-5)
(-4)
with one root.
Vertex =(2,0)
Sandy’s
Equation with
one claim
root.
(-3)
(-2)
(-1)
0
1
2
3
4
5
One Root
(-5)
(-4)
(-3)
(-2)
(-1)
0
1
2
3
4
5
Create a table of values
x
𝟑
𝒚=
𝒙
.25
.50
.65
Use the online calculator to find y
Create a table of values
x
.25
.50
.65
𝟑
𝒚=
𝒙
6
4.24
3.72
You can also use the graphing utility in the online calculator
To get a sense of what the graph should look like.
Choose Expression (y=)
Type formula here
Choose Graph
𝑐𝑜𝑠 =
𝑎𝑑𝑗
ℎ𝑦𝑝
𝑠𝑖𝑛 =
𝑜𝑝𝑝
ℎ𝑦𝑝
𝑐𝑜𝑠 =
𝑎𝑑𝑗
ℎ𝑦𝑝
=
𝑠𝑖𝑛 =
𝑜𝑝𝑝
ℎ𝑦𝑝
You’re being asked to
compare trigonometric ratios
When will cos equal sin?
When adj equals opp!
Isosceles
45-45-90
Triangle
opp
adj
A
5∘
55∘
25∘
35∘
65∘
75∘
45∘
15∘
85∘
𝑐𝑜𝑠 =
𝑎𝑑𝑗
ℎ𝑦𝑝
When is cos less than sin?
When opp less than adj!
≤ 𝑠𝑖𝑛 =
𝑜𝑝𝑝
ℎ𝑦𝑝
45∘
When the angle
Is less than 45∘
opp
A
adj
5∘
55∘
25∘
35∘
65∘
15∘
75∘
85∘
𝑐𝑜𝑠 =
When is cos less than sin?
When opp less than adj!
𝑎𝑑𝑗
ℎ𝑦𝑝
5∘
≤ 𝑠𝑖𝑛 =
𝑜𝑝𝑝
ℎ𝑦𝑝
45∘
25∘
When the angle
Is less than 45∘
opp
A
adj
55∘
35∘
65∘
15∘
75∘
85∘
𝑐𝑜𝑠 =
When is cos more than sin?
When opp more than adj!
𝑎𝑑𝑗
ℎ𝑦𝑝
5∘
≥ 𝑠𝑖𝑛 =
𝑜𝑝𝑝
ℎ𝑦𝑝
45∘
15∘
25∘
When the angle
Is more than 45∘
opp
adj
A
35∘
55∘
65∘
75∘
85∘
𝑐𝑜𝑠 =
When is cos more than sin?
When opp more than adj!
When the angle
Is more than 45∘
opp
adj
A
𝑎𝑑𝑗
ℎ𝑦𝑝
5∘
≥ 𝑠𝑖𝑛 =
𝑜𝑝𝑝
ℎ𝑦𝑝
45∘
15∘
25∘
75∘
35∘
85∘
55∘
65∘
-5
-4
+
3𝑥 + 1 = 3𝑥 − 3
-3
-2
-1
0
1
2
3
4
1 = −3
5
𝒂𝒙 + 𝒃
+
𝒂𝒙 + 𝒃
= = 𝒂𝒙𝟑𝒙
+−
𝒃3
3𝑥 + 1
2
3𝑥 + 1 =
3𝑥 − 3
2
= = 𝒂𝒙𝟑𝒙
+ 𝒃+ 1
3𝑥 + 1
2
3𝑥 + 1 =
3𝑥 + 1
3𝑥 + 1 = 3𝑥 + 1
1=1
☺
2
-5
-4
-3
-2
-1
0
1
2
3
4
5
3
-3
-5
-4
3𝑥 + 1
=
𝟑(2) + 1
=
𝒂𝒙 + 𝒃
𝟏(2) + 𝒃
7 = 2+𝑏
2
7 = 2+𝑏
7=2+b
5=b
2
-3
-2
-1
0
1
2
3
4
5
||
|
11
6
+ 7 + 6 = 19
The critical point of the graph
Of 𝑥 is A(0,0)
A(0,0)
x
𝒙
0
0 =0
4
4 =2
For 𝑥 𝑡𝑜 𝑏𝑒 𝑎 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
x must be ≥ 0
So, the domain of 𝑥
must be x ≥ 0
B(4,2)
The critical point of the graph
Of 𝑥 − 2 + 3 is A’(2,3)
A(0,0)
x
𝒙
0
0 =0
4
4 =2
x
B(4,2)
𝒙−𝟐+𝟑
2
2−2+3=𝟑
For 𝑥 − 2 𝑡𝑜 𝑏𝑒 𝑎 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
x-2 must be ≥ 0
So, the domain of 𝑥 − 2
must be x ≥ 2
This is a translation of 2 to the RIGHT
and UP 3
A’(2,3)
A(0,0)
x
𝒙
0
0 =0
4
4 =2
x
B’(6,5)
B(4,2)
𝒙−𝟐+𝟑
2
6
2−2+3=𝟑
6 − 2 + 3 = 5 Perform the same translation for B
start
end
50%
50% -34% =
Standard Deviation is the
average distance from each value
to the mean (average)
About 68% of data falls within one
Standard Deviation of the mean.
Only 16% of data falls below one
Standard Deviation (x<62).
16%
68%
34%
-3
34%
+3
In a Normal Distribution
The mean and median are equal
So 50% of the data is below the mean.
50%
16%
About 68% of data falls within one
Standard Deviation of the mean.
Only 16% of data falls below one
Standard Deviation (x<62).
Click between
62 and 65
16%
50%
In a Normal Distribution
The mean and median are equal
So 50% of the data is below the mean.
a
b
=c
= 14
𝑎 + 𝑏 + 130=
14 + 130
𝑎2 + 𝑏2 = 130
+𝑏𝑏 + 130=
= 16 + 130
𝑎𝑎+
𝑎2 + 𝑏2 = 130
a
x
=c
= 14
𝑎 + 𝑏 + 130=
14 + 130
𝑎2 + 𝑏2 = 130
+𝑏𝑏 + 130=
= 16 + 130
𝑎𝑎+
𝑎2 + 𝑏2 = 130
𝒙𝟐
2
4
3
9
4
16
5
25
6
36
7
49
8
64
9
81
10
100
11
121
b
We need two numbers
Whose squares add to 130
49 + 81 = 130
7 + 9 = 16
9 + 121 = 130
3 + 11 = 14
a
b
3
=c
a=3 b=11
7 + 9 = 16
11
7
𝟏𝟑𝟎
9
𝟏𝟑𝟎
A thrown ball follows a parabola.
Since it goes up first and then down,
It must open down
So, is in the form −𝒙𝟐 + 𝒃𝒙 + 𝒄
Initial height < 4.5
Maximum height will be
at the vertex.
x value of vertex =
−2
𝑥 = 2(−1) =1
−𝑏
2𝑎
𝑦=− 1
2
+ 2 1 + 3 =4
Maximum height will be
at the vertex.
x value of vertex =
𝑥=
−5
2(−1)
5
=2
−𝑏
2𝑎
5
𝑦=−
2
2
5
1
25
25
=9
=−
+
+
3
+5
+3
4
4
2
2
Rectangular Prism
V = lwh
V = x•x•(x+3)
V = 𝑥 •(x+3)
2
V = 𝑥 3 +3𝑥 2
𝒙𝟑 +3𝒙𝟐
Sphere
𝑥
r =2
4
V = 3 𝑟3π
4 𝑥
𝑉=
3 2
𝒙𝟑 +3𝒙𝟐
3
π
4 𝑥3
• π =
3 8
π𝑥 3
6
4𝑥 3 π
24
=
𝑥3π
6
Y-intercept is where x = 0
𝑓 0 = 500 1.015 0
𝑔 0 = 500 1.021
0
Anything to the zero power = 1
𝑓 0 = 500(1)=500
𝑓 1 = 500 1.015
𝑓 1 = 507.5
1
<
𝑔 0 = 500(1) = 500
𝑔 1 = 500 1.021
𝑔 1 = 510.5
1
An exponential function
is in the form: f(x) =a𝑏 𝑥 +k
The horizontal line
Y=k is the asymptote,
A line the graph approaches
but never reaches.
Y=0 {x-axis}
f(x) =a𝑏 𝑥 +0
f(x) =a𝑏 𝑥
An exponential function
is in the form: f(x) =a𝑏 𝑥 +k =a𝑏
=1𝑏𝑥𝑥+0
+0
(0,1)
1=a𝑏0
f(x) =𝑏 𝑥
(0,1) is a point on the graph
Y=0 {x-axis}
1=a(1)
f(x) =a𝑏 𝑥
An exponential function
is in the form: f(x) =a𝑏 𝑥 +k =a𝑏
=1𝑏𝑥𝑥+0
+0
f(x) =𝟐𝒙
2=𝑏1 so, b=2
f(x) =𝑏 𝑥
(1,2)
(1,2) is a point on the graph
f(x) =𝟐𝒙
f(6) =𝟐𝟔
f(6) =64