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MIT Academy Period:Algebra 2 Name:
Chapter 6 Assessment Complete these problem to make sure that you are prepared for the test.
1. Ifj(x) = 2x3+7 and g(x) = -3(x -9), find the value of each expression below.
= -5~j.t3
00 . 3,e
e.) Aref(x) and g(x) inverses of each other? Justi~’
W~
6x2—x—2
x2—4x—2V
if X -21
~.A~
a.)jt-3) ~ 2tt3)h43
1~ytc.)f(g(-3)) /
b.)g(-3)
t _3(’~~,,) ~427
ck€J
t±, ~~scra
~
your answer
c~~)) s~et-tdr’3 ._tj
— 4x + 3b.) x2—9
KCF~‘2 ‘j,( 43 _________
)C~~ ‘1
r
2. Simpli~’ the following expressions:
a)C-4~j XYXY8x3 4x2y
r
1C~
Af
4 ~2
~%2J1$13~(~~
•, ~t;
314-’ I(~A
~k{eVc
a
Algebra 2MIT Academy3. Subtract and simpli~’ the following expressions:
x—2 x—4a.) ~ — x —3
I x—1 x+Ib.) 2 ÷ x2+2x—3 — 2x+6
I (~Y,cj) —
jt3 2-26ck3) )6JY)2&.3) ____
_F~ I)tt3-42-Xj .1- -
= 26fl) ___
.
5. Graph on àometric graph paper
) .—i~
j..t; (2~≠t9) -3~
6~ 2t~2o$
Sx1,20A1& /
a. Sketch a graph of the solution to the equation: 4x ÷ 6y÷ 3z = 24.~.•.•.•.•.•
• . • . . . • -lit
•:•:•.•.•:• ••:•:•:•:•:~:•:•:•:~:• :•:•:•:•:•: .t.~t3±.!~°• : • : • : : : : : • : •
:~:•:~ :~: •: :• :~ :• : • : • : • : • : • : • : • : • ,~ y:rf.C:.:::.:::t~__~~___
• ~
-3r-2’l(ero-Ø)-
b. Graph the point (2,3,5). What are three otherpoints that have the safiuiisometric d~t position?
6, ‘.~)(C) L1 i)
.
Itzt ~,
4,2(’9~)
Name:Period:
~tCD
~2 4j_15
4-~c’ -20) ~&t3)
4. Identify the vertex and thejine of s mmetry: f(x) = 5x2 + 20x — 16
~
~Øj~
•p 4’):
*J .r~ yecies: (—2, -36)lute 0ç 7$4M4~O( =-
.
• •
U,,
.1 4~ €~( -$~
lest/toe ,h0~,A’
Nc4ke~ s.J4~ty ~? 2/ Name:/ Period:
!~?jfl: iC
7. Solve each equation below for x
_1 tq,,)~3b~s:2 t—(~43b5
j —23 t 3b
Algebra2 LØLIA?? -4, Vnplruc’<Mil’ Academy6. Solve for x, y, an~he system of equations at below.
— 2y + 3z = 32x+y+5z=8
3x — y — 3z =
3cc_L?~
/AU ~1~4d~oi~ -‘
cnce( ~_1. v’ar~We
Sc)”
c~x~4 iq~g
~3zr~fl
~5’ ~27t44
5x.2t .~ -.-f4L
..t ~3~z.tfl
_IIz:33
42(3):: -pi
:a=3
(-‘i) 28
4~~
.gflz3
.34)23
= 3ge3a. 18 = 2(3)(x+3) b. 128 = 43X
3$
I, 21 ~~7KI’
7~C≠ ~t,L__ ~~73
pg
by’ P6
06
_3P
•, ~El.-34 —c’8 What f~ the equation of the parabola passing through the points (3, 1), (2, 3), and (0, —5)?
~KQ1ACA
~‘$ ~~ezy~sbXtC
~~4’~3~)4c ~_,J~7~~3b 4C IrTa43b-5 —v
—? e=-S j1Rt3b~,a4~,YtbS)1C _73zt4~÷2b-fC —(2 ~a
42 —in-—-p 12-ia..’
“—p k42b3.2~4Q)
34
—1 —t—P —D 2xr •13
11. Find the inverse of each of the functions below. Write your answers in function notation.
a)p(x)~(x+4)2+1
2)(~
2(y-O ~4q)
I,
X~ 34~d7’’
-2
&
Algebra 2 Name:MIT Academy Period:9. In parts (a) through (b), rewrite each expression as a single logarithm. In parts (c) through (f), solve each equation.
a. log2(30x)-1 log2(6) ..rtt,, 4 (2$~) —si
b. 21o~(x)+log3(5) ~, hy3X’~~;~ ~$36x2
c. log7(3x—2)2 —.D 72~3(~2 -~ qy~ 3~-2 —) hg9 3~ —t
d. Iog(2x+1)=—1 —a ~
e. log5(3y) + logs(9) = logs(405) —‘ 275 2?y 4;#cfl ‘~ç4~ =4O~ ~-, ~, :15
f. log(x)+log(x+21)2 —v 2 —w ~ rP
~ ~
10. Find the equation of the exponential functi~asses through the points (2,264) and (6, 4044) and has a horizontal
uYtOl. ,
asymptote at y = 12. (hint: think about 6.2.3 and use y = ab’ + c.
C:)2
2≤~ta.h~t12
a52:ct~b’ 1%
252rO~ ___
r tjO3Jt26) W~’z52
(Ezr1~
.
b.) k(x) = 31og2(x —2)
I-It2)
y
MIT Academy Period:Algebra 2 Name:
12. Define the vocabulary. You may use words, examples, pictures, symbols to convey what you mean.
Logarithm:
Power property of logs:
Asymptote:
X-intercepts:
Plane (in the context of algebra 2):
yc-(.6 Asd ycif’
b eRi~ catet *i~oo+ < - ~
e. Solve and show the solution on a number line: 1x + 3 <x2 — 2x — 3.
)E:~t7 e7.5,2~)t_d
a line, one is a parabola {extra credit if you complete the
y
‘1,5)
square to put it into y = a(x — h)2 + k})13. Graph the system: (hint: one of them is(‘0,3)
y=~x+3 ~kpe~fr~°~
y = x2 — 2x —3
I
b. Use the graph to solve + 3 = x2 — 2x —3.
rtJ
O ~
O’
t@7 (~)-3
-3-I
~o)
(~,-3)
~c.Solvey = __________
d. What is the difference between the answer to$andt Explain why this is so.
liD’,
4frre ;~,4ke
—‘.5
$t3
‘Vbeio~o44e b/k~4 01g~
Ar% y4-15
Name:Algebra 2MIT Academy Period:
For problems f and g, solve by shading the appropriate region.
i.y≤~x+3
y ≤ x2 — 2x — 3
j.y≥~x+3 k.y≥~x±3
y≥x2—2x—~ y≤x2—2x—3
(/7.it/ /~c //7
Jl////li ~
~,4 7,/f
h. y — 2x —3
y ≥ + 3 a6qre
.
.42w1Q ce
S.
/1’
/I’
/
/L