Resources

Resources

Transformations of absolute value functions

Choose the correct symbol to compare the expressions. Do not multiply.

\newline

8 \times \frac{1}{1} ? 8

\newline

\newline

>

\newline

<

\newline

=

Colin noticed it rained

\frac{7}{8}

of the days that he recorded the weather in Jackson County. In Putnam County, Hector found that it rained

\frac{1}{6}

of the days he kept records. According to the records, which county had rain a greater fraction of the time?

\newline

\newline

(A)Jackson County

\newline

(B)Putnam County

This afternoon, Richard got through

\frac{5}{9}

of the pages he was hoping to read. Meanwhile, Dave read

\frac{2}{5}

of the pages he had planned to read. Who got through a greater fraction of his reading?

\newline

\newline

\newline

\newline

p(x)=5x^{3}-44x^{2}+61x+14

has a known factor of

\newline

(x-7)

\newline

\newline

p(x)

as a product of linear factors.

\newline

p(x)=

Dingane has been observing a certain stock for the last few years and he sees that it can be modeled as a function

S(t)

t

(in days) using a sinusoidal expression of the form

a \cdot \sin (b \cdot t)+d

\newline

t=0

, the stock is at its average value of

\$ 3.47

91

25

days later, its value is down to its minimum of

\$ 1.97

\newline

S(t)

\newline

t

should be in radians.

\newline

S(t)=

h(x) = 2^x

change over the interval from

x = 5

x = 6

\newline

\newline

h(x)

100\%

\newline

h(x)

2\%

\newline

h(x)

2

\newline

h(x)

2\%

h(x) = 2^x

change over the interval from

x = 1

x = 2

\newline

\newline

h(x)

2\%

\newline

h(x)

2

\newline

h(x)

100\%

\newline

h(x)

2\%

g(x)

4

over every unit interval in

x

g(0) = 0

. Which could be a function rule for

g(x)

\newline

\newline

g(x) = -4x

\newline

g(x) = 4^x

\newline

g(x) = 4x

\newline

g(x)

-x

4

g

be a function such that

g(4)=8

g^{\prime}(4)=-3

\newline

h

be the function

h(x)=\sqrt{x}

\newline

H

be a function defined as

H(x)=g(x) \cdot h(x)

\newline

H^{\prime}(4)=

\bar{a}= \hat{i}+2\hat{j}-\hat{k}

\bar{b}= \hat{i}+\hat{j}-\hat{k}

be two vectors. If

\bar{c}

is a vector such that

\bar{b}\times \bar{c}= \bar{b}\times \bar{a}

\bar{c}\cdot \bar{a}=0

\bar{c}\cdot \bar{b}

20

years younger than Ishaan. Ishaan and Christopher first met two years ago. Fourteen years ago, Ishaan was

3

times as old as Christopher.

\newline

How old is Christopher now?

\newline

\square

p(x)=x^{3}-6 x^{2}+32

has a known factor of

(x-4)

\newline

p(x)

as a product of linear factors.

\newline

p(x) = \square

Why does multiplying by

10

shift each digit one place to the left?

\newline

(A) Each digit represents

10

\newline

(B) Each digit represents

10

\newline

(C) Each digit represents

10

times its original value.

\newline

(D) Each digit represents

\frac{1}{10}

of its original value.

(b) The first snow of the season begins to fall during the night. The depth of the snow,

h

, increases at a constant rate through the night and the following day. At

6

am a snow plough begins to clear the road of snow. The speed,

v \, \text{km/h}

, of the snow plough is inversely proportional to the depth of snow. (This means

v=\frac{A}{h}

A

is a constant.) Let

x \, \text{km}

be the distance the snow plough has cleared and let

t

be the time in hours from the beginning of the snowfall. Let

t=T

6

am. (i) Explain carefully why, for

t \geq T

6

0

6

1

is a constant. (ii) In the period from

6

6

3

am the snow plough clears

6

4

of road, but it takes a further

6

5

hours to clear the next kilometre. At what time did it begin snowing?

Consider the expression

\newline

(\frac{1}{2})(2a+1)(a+3)

\newline

2

descriptions of the parts of the expression.

\newline

The entire expression is the product of

\square

\newline

(2a+1)

2

\square

f(x) = 2x^5 + x^4 - 18x^3 - 17x^2 + 20x + 12

x - 3

f

, what is the value of

f(3)

f(x)=-3 \sec (2 x)

f^{\prime}(x)

\newline

f^{\prime}(x)=

x^{3}+4-5 x^{2}=-5 y+y^{2}

\frac{d y}{d x}

(1,5)

\newline

\left.\frac{d y}{d x}\right|_{(1,5)}=

-5 y+3 y^{2}+4+5 x^{3}=x^{2}

\frac{d y}{d x}

(-1,2)

\newline

\left.\frac{d y}{d x}\right|_{(-1,2)}=

-4-y+x^{2}=y^{2}+3 y^{3}

\frac{d y}{d x}

(-3,1)

\newline

\left.\frac{d y}{d x}\right|_{(-3,1)}=

-5-x^{3}-y^{3}=-3 x^{2}

\frac{d y}{d x}

(2,-1)

\newline

\left.\frac{d y}{d x}\right|_{(2,-1)}=

f(x)

g(x)

are differentiable. The function

h(x)

\newline

h(x) = g(x) - f(x)

\newline

f(2) = 5

f'(2) = -1

g(2) = -7

g'(2) = 3

h'(2)

\newline

Simplify any fractions.

\newline

h'(2) =

f

be a function such that

f(9)=-54

f^{\prime}(9)=6

\newline

h

be the function

h(x)=\sqrt{x}

\newline

\frac{d}{d x}\left[\frac{f(x)}{h(x)}\right]

x=9

g(x)

g(x)

is the translation

1

f(x) = |x|

\newline

Write your answer in the form

a|x - h| + k

a

h

k

\newline

g(x) =

\newline