mhf4u_unit1_assessment_part1
pdf
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School
York University *
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Course
MHF4UA
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
4
Uploaded by DeanRainOctopus21
Copyright © 20
21
The Ontario Educational Communications Authority. All rights reserved.
MHF4U Learning Activity 1.11
Unit 1 Assessment, Part 1 TVO ILC
1 Unit 1 Assessment, Part 1 1. Fill in the tables below for each of the following polynomials: [K10] _
a) f
(
x
)
= −
1
4
x 3
− 13
x 2
+ 3
x − 9 Degree Type of polynomial Sign of leading coefficient
End behaviours Domain b) f
(
x
)
= 6
x
4 + 7
x
2 − 8
x + 9 Degree Type of polynomial Sign of leading coefficient
End behaviours Domain 3
Cubic
Negative
x-> -
∞, y->∞
x-> ∞ , y->-∞
Domain is not restricted
Domain is not restricted
4
Quartic
Positive
x-> -
∞, y->∞ x->
∞ ,y->∞
Copyright © 20
21
The Ontario Educational Communications Authority. All rights reserved.
MHF4U Learning Activity 1.11
Unit 1 Assessment, Part 1 TVO ILC
2 2. Fill in the table below for the polynomial function: [K9]
f
(
x
)
= −
(
x
− 4
)
(
x
+ 1
)
2
(
x
− 5
) x
– intercepts
and behaviour of graph at each x
−
intercept
y
− intercept
End behaviours Sketch: At x=4 Graph raises up through x axis
At x=-1 Graph bounces on the x axis
At x=5 Graph decreases down through x axis
20
x ->-
∞, y->∞
x->
∞, y->-∞
x
y
1
5
-5
-5
-1
5
-1
Copyright © 20
21
The Ontario Educational Communications Authority. All rights reserved.
MHF4U Learning Activity 1.11
Unit 1 Assessment, Part 1 TVO ILC
3 3. Solve the following equation algebraically. Explain your process. [A6], [C3]
0 = 18 x
4
+ 87 x 3
+ 3 x 2
− 108
x 4. a) Describe the transformations on the cubic function below.
Explain your thinking process. [C4]
_
f
(
x
)
= 7 [
1
(
x
− 5
)
]
3
− 45 2 _
b) The point (-9, -2446) is on the transformed function f
(
x
)
= 7 [
1
(
x
− 5
)
]
3
− 45
use the
2 transformation statement the find the original point on the parent function. Justify your
answer. [T3], [C2] x=1
0 = 3x(6 x^3 + 29 x^2 + x − 36
6(1)^3 +29(1)^2 + 1 - 36 = 0
=35+(-35) = 0
x= -(35) +/- √35^2 - 4(6)(36)
2(6)
x= -(35) +/- √1225 - 864
12
x= -35 +/- 19
12
x= -9/2 , -4/3
x= 0,1,-/3, -4/3
x-1
√6 x^3 + 29 x^2 + x − 36
6x^2 + 35x +36
- 6x^3 - 6x^2
35x^2 +x
- 35x^2 - 35x
36x-36
- 36x-36
0
a = 7 means vertical stretch by 7 since a>1
k= 1/2 means a horizontal stretch by 2 since k<1
d= 5, we get d by setting x-5=0 then isolating x. Func moves 5 units right since d is positive
c= -45, 45 units down since c is negative
(x,y) -> (x/1/2 - 5, 7y - 45)
(x,y) -> (-9, -2446)
1/2
x
- 5 = -9
x
1/2
= -9 + 5
x = -2
x
1/2
= -4
7y - 45 = -2446
7y= -2446 + 45 7y = -2401
y = -343
original point on parent
function is (-2, -343)
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Copyright © 20
21
The Ontario Educational Communications Authority. All rights reserved.
MHF4U Learning Activity 1.11
Unit 1 Assessment, Part 1 TVO ILC
4
5. Determine the values of m
and n for f
(
x
)
= mx
3
+ 20x
2
+ nx− 35
given that
(
x
+ 1
) gives a remainder of zero, and when divided by (
x
− 2
)
the remainder is 45. [T5],
[C3]
6. Use algebra to determine whether the following function is even, odd or neither. [A6]
a)
f
(
x
)
= 55 x
4
− x
2
− 78 b)
f
(
x
)
=37
x
3
+93
x
f(2) = m(2)^3 +20(2)^2 + n(2)-35
45= 8m +80+2n - 35
45-45 = 8m +2n
0= 8m+2n
f(-1) = m(-1)^3 +20 (-1)^2 + n(-1) -35
0= -m + 20(1) - n - (35)
0= -m-n-15
15= -m-n
m= -15-n
8(-15-n) +2n=0
-120-8n+2n=0
-120-6n=0
-120=6n
n=-20
8m+2(-20)=0
8m-40=0
8m=40
m=5
f(-x)=55(-x)^4 - (-x)^2 - 78
=55x^4 - x - 78
-f(x)=-(55x^4 -x^2 -78)
= -55^4 +x^2 +78
f(-x)=f(x), Function is even
f(-x)=37(-x)^3 +93(-x)
= -37x^3 - 93x
-f(x)= -(37x^3 + 93x)
=-37x^3 - 93x
f(-x) = -f(x) , Function is odd