Pre - Calculus 12
Unit 1 Review
1.1-1.3 - Transformations
Big ideas:
 A function y = f(x) can be transformed by: fill in the blanks
o Vertical reflections (reflections in x-axis)
–y = f(x) or y   f ( x)
o
Horizontal reflections (reflections in y-axis)
y  f (  x)
o
Vertical stretches of a
y = af(x) or
o
Horizontal stretches of b
y  f ( b1 x)
o
Vertical translations of k
y-k = f(x) or y  f ( x)  k
o
Horizontal translations of h
y  f ( x  h)
o
Putting it all together, we get
y  af
1
a
( y )  f ( x)
  x  h   k
1
b
or
1
a
( y  k )  f ( b1 ( x  h))

We can transform a single (x,y) location – or the entire graph – by creating a mapping rule.

The mapping rule for


Things “inside the function” are “always opposite” because they are transformations to get back to original.
When looking at a graph and identifying the transformations that have been applied to it, we should
o First identify the reflections
o Then the stretches
o Then look for the translations
o Invariant points, particularly the origin, are useful.
1
a
 y  k   f  1b  x  h   is ( x, y)  (bx  h, ay  k )
1. Construct a table of values for the base functions (the functions to be transformed) for:
a) y 
x
X
y
0
0
1
1
4
2
9
3
16
4
b) y  x 2
X
y
-3
9
-2
4
-1
1
0
0
1
1
2
4
3
9
c) y  x
X
y
-3
3
-2
2
-1
1
0
0
1
1
2
2
3
3
Think of these ordered pairs when
you are transforming graphs! Will
help you recognize stretches.
2. Transform y = f(x) (at right) and plot the image function for each:
Points: A(-3, 3) B(0, -1) C(1.25, 1) D(2.75, -3) E(4, -1) F(5, -1)
A
C
a) VS = ¼, Reflection in x-axis, VT = 2, HT = -1
E
Use the mapping rule ( x, y)  ( x  1,  y  2)
1
4
New Points: (-4, 1.25) (-1, 2.25) (0.25, 1.75) (1.75, 2.75) (3, 2.25) (4, 2.25)
b) 0.5( y 1)  f ( x  3)  0.5( y  1)  f (( x  3))
Mapping Rule: ( x, y)  ( x  3, 2 y  1)
New Points: (6, 7) (3, -1) (1.75, 3) (0.25, -5) (-1, -1) (-2, -1)
B
F
D
c) The highest to lowest points is now 3 units, the most left to most right points is now 12 units, there have been
no reflections, and (0,-1) is now mapped to (-2,-4)
Highest to lowest used to be 6, now 3, VS  63  12 , width used to be 8, now 12, HS  128 
3
2
HS will not affect the zero so HT of -2. Apply VT of ½ to -1 to determine what the VT will have to be -3.5.
Mapping Rule: ( x, y)  ( 32 x  2, 12 y  3.5)
New Points: (-6.5, -2) (-2, -4) (-0.125, -3) (2.125, -5) (4, -4) (5.5, -4)
3. In your textbook. Page 56 #4 and #8
4. 4. Page 57 #9-12
Answers in Text
5. In the graph at right, identify the transformations
that map f to g.
(f is the original).
f
VR (reflection in x axis)
HS 
new width 3.5 1


old width
7 2
No VS
HT 2.5 and VT 1
g
Mapping Rule: ( x, y)  ( x  2.5,  y  1)
1
2
1.4 - Inverses
Big ideas
o Inverse functions always undo each other.
o The input of one function becomes the output of the other function, and vice versa
o We can find the equation of an inverse function by interchanging x and y and then solving for y.
o The graphs of inverse functions are reflections of each other across line y = x.
o The domain of the function is the range of the inverse and vice versa
o Not all functions can have inverses - because the both a function and its inverse need to be functions.
o We can make a function have an inverse if we restrict its domain.
1. What is the inverse of y = x3? Is this function always the inverse?
4
x  y3
3
y
3
2
1
x  y or x 3  y
1
x
Always a function – original graph passed the horizontal line test.
−6
−4
−2
2
−1
−2
−3
−4
4
6
2. If g ( x)  x 2  4 x and x  2 , find g 1 (5)
g ( x)  x 2  4 x  4  4
g 1 (5)  5  4  2
g ( x)  ( x  2) 2  4
Inverse
g 1 (5)  1
Or you could determine where g(x) was
equal to 5 and use the answer that satisfied
the domain. Think about the idea of
inverses, if we let y = 5 in the original and
solved for x then those x values will be the y
value for the inverse.
x  ( y  2) 2  4
x  4  ( y  2) 2
x4 2 y
3. Sketch the inverse of the function at right.
Same sketch as on ICA – you better be able to graph the inverse!
4. Algebraically, determine the inverse to:
a) x  7 y  11
b) x  3 y  1  4
c) y  3( x 2  6 x  9  9)  4
x  11
 y 1
7
x  4  3 y 1
y  3( x  3) 2  23  x  3
( x  4) 2  3 y  1
Inverse
( x  4) 2  1  3 y
1
3
( x  4)   y
2
1
3
1
x  3( y  3) 2  23
1
3
( x  23)  ( y  3) 2
1
3
( x  23)  3  y 1
d)y 
1
1
x
1
1
y
1
x 1 
y
1
y
x 1
x
s 1 ( x) 
1
f) typing error
x 1
1
x
y 1
1
y 1 
x
1
y  1
x
1
p 1 ( x)   1, x  0
x
e) y 
g) y 
1
x2
1
, x 1
x 1
1
y2
1
y2 
x
1
y
x
x
f 1 ( x ) 
1
,x0
x
The chapter test on pages 58-59 is also pretty decent as an overall summary of sections 1.1 – 1.4. There are
answers in the back of the book, of course.
2.1 – 2.2 Radical Functions with Linear and Quadratic Radicands
Big ideas
o The most important thing to establish with a radical function is its domain. This tells most of the story.
o The graph of a radical function with a linear radicand takes the shape of a horizontal parabola.
o We can understand radical functions with linear radicands by figuring out 1) domain, 2) stretch,
3) reflection
Or we can understand them by applying transformations to the base function y  x
When finding the equation of a horizontal parabola from a graph, we can choose either a horizontal
stretch or a vertical; but we don’t need both.
If a radical function has a quadratic radicand,
o
o
o
x 2 equal? y  x
o
What does
o
When trying to understand a radical function with a quadratic radicand, it is very useful to sketch a graph
of its radicand, because it can help you understand the domain of the original function.
The domain might be such that x can get very, very large, and if it does, then the original function
o
approaches the curve, y  x
o
If x can’t get very, very large, then then original function takes the shape of an ellipse.
1. You’re in physics class, and you’re experimenting with simple pendulums. Greer has told you that the length of time it
takes for the pendulum to make one full swing (T, in seconds), is described by: T  2
L
9.8
, where L is the length of the
pendulum’s string, in metres. Paulie says, “sweet! I’m going to double the length of my string. That’ll mean that the time
taken for one swing will be twice as long!” Is he right or wrong? Why?
Answer: The time will become 2 times as long. L is underneath the radical sign – multiplying it by two puts the two
under the radical sign as well, so 2 .
2. The dimensions of a rectangular prism are: width = a¸ depth = b, height = c. What is the
length of the diagonal from one corner to the other (thru the middle of the box)?
Answer: This will involve using the Pythagorean Theorem twice to find the length of the diagonal.
First you will need to find the length of the diagonal (x) of the base of the box, a 2  b2  x 2 .
Then to find the diagonal through the middle of the box, x2  c 2  d 2 or a 2  b2  c2  d 2 .
This will give us, d  a 2  b2  c 2 . (You might remember this from math 10)
3. Sketch a graph of each, without using technology
a) f  x   2 x  4  3  f ( x)  2( x  2)  3
Need to separate stretch and translation!
Mapping Rule: ( x, y)  ( 12 x  2, y  3) Apply to original points or start at (2, -3)
2
and go over ½ and up one, over 2 and up 2 and over 4.5 and up 3…..
1
y
x
5
−1
−2
−3
−4
10
b) f  x     x  3  f ( x)   ( x  3)
2
y
Need to separate the HS from the
1
translation!
x
Mapping Rule: ( x, y)  ( x  3, y)
Endpoint is (3, 0) with a horizontal reflection
(over the y axis), a vertical reflection (over the x axis), and no stretches.
−4
−3
−2
−1
1
2
3
4
−1
−2
−3
−4
c) The inverse of y  2 x 2  6 x  7 , x 
3
2
Need to complete the square!
y  2( x 2  3x  94  94 )  7
8
y  2( x  32 ) 2  92  7
7
y  2( x  32 ) 2  52 , x 
6
3
2
5
4
Inverse
x  2( y  ) 
3 2
2
2
1
(x  )  ( y  )
3 2
2
5
2
1
2
(x  ) 
5
2
3
2
This is the inverse!
3
5
2
x  52  2( y  23 ) 2
1
2
y
−1
5
2
x
1
2
3
4
5
6
−1
3
2
Endpo int ( , )
−2
4. Find the equation of both the upper and lower horizontal parabolas, at right.
Vertex (4, -3) VS of 3 (over one and up three – should be over one and up one)
Upper: y  3 x  4  3
Using the equation:
y  a x  4  3 Pt (5, 0)
0  a 54 3
Lower: y  3 x  4  3
3a 1
3a
5. Find the equation of half a horizontal parabola with a starting point at (2,3) and passing through (-5,-9).
Couple of ways you can do this, but sketch the 2 points so you can see if there are reflections or not. Remember the first
point should be over one and down one from the endpoint…..
So VR, HR, VT of 2, HT of 3, VS of 12, HS of 7
y  12  17 ( x  2)  3
You could also use equation but it will just solve for a
vertical stretch and it will be a nasty one.
y  a ( x  2)  3 pt (5, 9)
a
12 12 7

7
7
1/7 out of radical!
6. Sketch a graph of each, without using technology. Be sure to state the domain (which will help you with your sketch).
You will need to label a few key points!
f  x   12  3 x 2
f  x   x 2  25
f  x   7 x 2  12
12  2 x 2  0
x 2  25  0
2
a) 7 x  12  0
b) x 2  25
x 2   127 always true
c)
x 5
xR
10
y
6  x2
6 x
x  5 or x  5
20
12  2 x 2
 6x 6
y
15
y
5
x
15
−10
−5
5
10
10
−5
10
5
−10
5
x
−15
x
−3
−2
−1
1
2
−3
−20
−2
−1
1
2
3
3
−25
−5
7. Consider the general function y  ax 2  c . We know that this function will take the shape of a hyperbola in one
piece, a hyperbola in two pieces, or an ellipse. Describe the conditions for each of these three shapes to occur. I.e. “if a
is ……. and c is ……, then we get the graph of a ……………”
Look at question #6
a) This is the shape of a hyperbola in one piece: positive a, positive c.
b) This is a hyperbola: positive a, negative c.
c) This is an ellipse: negative a, positive c.
Do you know why we do not consider the case with a negative a, negative c?
2.3 Solving Radical Equations
Big ideas
o We can solving a radical equation algebraically or graphically.
o What’s the song we sing that describes the steps in solving a radical equation algebraically? Isolate the radical,
square both sides and check for extraneous roots…..(not sure if that is what Woods sings, but it is what I say)
o What is the danger in using the algebraic method? Extraneous roots.
o How do we deal with this? What’s the technique? Check each root in the original equation to see if they work.
o If we want to solve graphically, then we try to find the intersection of the entire function with the x axis, or find
the intersection of the left side and the right side of the equation.
1. Solve each equation algebraically, making sure to check your solutions for validity.
a)
8  x  x  3
8  x  x  3
11  2 x
 112  x extraneous
x  3  2 x2  7
x2  6x  9  2 x2  7
b) 0  x 2  6 x  16
0  ( x  8)( x  2)
x  8 x  2 both good
2. Solve each equation algebraically, checking your solution with a graphing tool like DESMOS or equivalent
x2  2  5  0
4  x2 x7
x  3  64  x 2
x2  2  5
2
a) x  2  25
x 2  23
x   23
b)
x  3  64  x 2
x 2  x  67  0 (QF )
x  7.70 x  8.70(ext )
4 x  2 x7
c) 16  8 x  x 2  4( x  7)
x 2  12 x  44  0
no solution
both good
3 x 2  11  x  1
3 x 2  11  x 2  2 x  1
2
d) 2 x  2 x  12  0
2( x 2  x  6)  0
2( x  3)( x  2)  0
x  3 x  2(ext )
3. Do the curves y = x + 3 and y  9  2 x 2 cross? If so, where?
x  3  9  2x2
x2  6x  9  9  2 x2
3x 2  6 x  0
These answers are both good! So the cross twice!
3 x( x  2)  0
x  0 x  2
4. The Beaufort Scale was designed to measure wind speed. The scale has a minimum value of 0 and a maximum value of
12, and approximated by B(v)  1.42 v  15  5.5 , where v is the wind speed in an open area at a height of
10 meters off the ground (v is measured in km/h).
a) Create a rough sketch of this function, being sure to establish a feasible domain. Let v  0 .
b) If the Beaufort Scale reads 0, what is the wind speed? Let B = 0 and solve for v, v = 0 km/h.
c) If the Beaufort Scale reads 12, what is the wind speed? Let B = 12 and solve for v, v = 136.88 km/h.
d) How much faster is the wind at a Beaufort number 6 than it is at 3? At 6 it is 50.58 km/h and at 3 it is 20.83 km/h.
e) Why would we bother creating a new scale for wind speed when we’ve already got a way to measure it (i.e. regular
speed)?
5. The chapter review and chapter test pages 99-103 are also excellent for 2.1 – 2.3, and there answers in the back of the
book.
LISTEN
If you want answers for sure, then do the textbook stuff first.
If you are asked to graph, then you can always check your work with DESMOS.
If you are asked to solve, then you can always check your solutions using WOLFRAM ALPHA. For example, if you
need to solve 2 x  5  x  6 , then go to WOLFRAM ALHPA and just enter (2x+5)^0.5 = x-6 and see what come up.
90% of this review could be done with DESMOS or WOLFRAM ALPHA.