radical functions class booklet solutions
DESCRIPTION
Topics 2 and 3 from our class handoutTRANSCRIPT
π(π) =πππ + π
Alberta Ed Learning Outcome: Graph and analyze radical functions. β’ Transformations of radical functions also includes sketching and analyzing the transformation of
π = π(π) to π = οΏ½π(π). The function π = π(π) should be limited to linear or quadratic functions.
Consider the coordinates of the 6 indicated points on the graph below. Do not label. Transform each point by the mapping rule (π₯,π¦) β (π₯,οΏ½π¦). Plot each new point. Sketch the resulting transformed graph.
A radical function can have the form π¦ = οΏ½π(π₯). In this topic weβll examine the characteristics of the graph of a radical function, along with the domain and range.
π(π) = ππ
Follow the same steps indicated in the task box for Explore 1.
State the equation of the transformed function
State the domain and range of both π¦ = π(π₯) and π¦ = οΏ½π(π₯)
Explain how you can derive the domain of a function
π¦ = οΏ½π(π₯), given the graph of equaton of π(π₯).
State the equation of the transformed function
State the domain and range of both π¦ = π(π₯) and π¦ = οΏ½π(π₯)
Explain how the graph of π¦ = οΏ½π(π₯) differs from the graph of π¦ = π₯.
Invariant Points (where π = π or π)
(0,2) (10,3)
(0,4)
(10,16) π = οΏ½πππ + π
π(π):
οΏ½π(π):
Domain: {π β πΉ}
Domain: {π β₯ βπ}
Range: {π β πΉ}
Range: {π β₯ π}
Excellent question! The function π = οΏ½π(π) is only defined where π(π) > π. So the domain can be found by locatnn the π-intercept of π = π(π) and determininn where the nraph is positiee (boie the π-axis)
π = οΏ½ππ
(π,ππ)
(π,π)
All points (π,π) β (π,οΏ½π)
π = οΏ½πππ + π
π(π):
οΏ½π(π):
Domain: {π β πΉ}
Domain: {π β πΉ}
Range: {π β₯ π}
Range: {π β₯ π}
Very similar β howeier unlike on the nraph of π(π), the nraph of οΏ½π(π) is always positiiee Since when x=0 we first SQU(RE the number, then square root it)
Given the graph or equation of a function π¦ = π(π₯), we
can obtain the graph of π¦ = οΏ½π(π₯) by transforming all
points (π₯,π¦) β (π₯,οΏ½π¦). See points ( and B
The domain of π¦ = οΏ½π(π₯) can be found by considering the zeros / π₯-intercepts of π¦ = π(π₯).
Since we canβt square root negatives, π¦ = οΏ½π(π₯) is defined wherever π(π₯) β₯ 0, that is, wherever the graph is above the π₯-axis. See point D
(D is the βstart pointβ for the domain)
π(π) = βππ + π
π = οΏ½π(π)
(βπ.π,π)
(βπ.π,π)
π¨π
π¨π π©π
π©π πͺ π«
The invariant points in the transformation from π¦ = π(π₯) to π¦ = οΏ½π(π₯) can be found by considering where the value of π(π₯) is 0 or 1. See points C and D.
(The square root of 0 is 0, and the square root of 1 is 1.)
Working from the equation of π¦ = β4 β 2π₯β¦ The domain is:
And the invariant points occur wherever we are square rooting 0 or 1. (β0 = 0 and β1 = 1)
1st invariant point is where π(π₯) = 0β¦ 2nd invariant point is where π(π₯) = 1β¦
Whatever we are square rooting cannot be negative. (That is, it must be β₯ 0)
4 β 2π₯ β₯ 0
β2π₯ β₯ β4
π β€ π
-2 -2 *When dividing (or multiplying) both sides of an inequality by 0, reverse the inequality direction!
4 β 2π₯ = 0
β2π₯ = β4
π = π
4 β 2π₯ = 1
β2π₯ = β3
π = π.π
Recall
π(π₯) is β4 β 2π₯β
So, coordinates of invariant point are (π,π ) So, coordinates of invariant point are (π.π,π )
1. For each given graph of π¦ = π(π₯), sketch the graph of π¦ = οΏ½π(π₯), and state its domain, range, and any invariant points.
(a)
π(π) = ππ β π
Domain of π¦ = π(π₯): Domain of π¦ = οΏ½π(π₯):
Range of π¦ = π(π₯): Range of π¦ = οΏ½π(π₯):
Invariant Points: π = βππ β π
(π,π)
(π,π)
{π β πΉ}
{π β πΉ}
{π β₯ π.π}
{π β₯ π}
Are on the graph of π(π) = ππ β π where the value (y-coordinate) is 0 or 1. POINTS are: and (1.5, 0) (2, 1)
(b)
π(π) = βπ.ππ + π
(c) π(π) = π.πππ β π
Domain of π¦ = π(π₯): Domain of π¦ = οΏ½π(π₯):
Range of π¦ = π(π₯): Range of π¦ = οΏ½π(π₯):
Invariant Points:
Domain of π¦ = π(π₯): Domain of π¦ = οΏ½π(π₯):
Range of π¦ = π(π₯): Range of π¦ = οΏ½π(π₯):
Invariant Points:
(d) π(π) = βππ + ππ
Domain of π¦ = π(π₯): Domain of π¦ = οΏ½π(π₯):
Range of π¦ = π(π₯): Range of π¦ = οΏ½π(π₯):
Invariant Points:
(e) π(π) = π.πππ + π
Domain of π¦ = π(π₯): Domain of π¦ = οΏ½π(π₯):
Range of π¦ = π(π₯): Range of π¦ = οΏ½π(π₯):
Invariant Points:
{π β πΉ}
{π β πΉ}
{π β€ ππ}
{π β₯ π}
POINTS are: and (10, 0) (8, 1)
{π β πΉ}
{π β πΉ} {π β€ βπ,ππ π β₯ π}
{π β₯ π}
Are on the graph of π(π) where the value (y-coordinate) is 0 or 1. POINTS are: and (-2.45, 1), (-2, 0), (2, 0) (2.45, 1)
π = ββπ.ππ + π
On the graph of π(π) the y-intercept is 5
β¦on π = οΏ½π(π) itβs βπ, or approximately 2.24
π = οΏ½π.πππ β π
DOM(IN: Graph of οΏ½π(π₯) is defined where 0.5π₯2 β 2 β₯ 0
(Solve graphically β what are the x-intercepts of π(π₯) = 0.5π₯2 β 2 / where is the graph above the x-axis?)
0.5π₯2 β 2 = 1 0.5π₯2 = 3
π₯ = Β±οΏ½3
0.5
Find π₯ where π(π₯) = 1
{π β πΉ}
{π β€ ππ} {βπ β€ π β€ π}
{π β€ π β€ π}
Are on the graph of π(π) where the value (y-coordinate) is 0 or 1. POINTS are: and (-3.87, 1), (-4, 0), (4, 0) (3.87, 1)
βπ₯2 + 16 = 1 15 = π₯2
π₯ = Β±β15
Find π₯ where π(π₯) = 1
π = οΏ½βππ + ππ
{π β πΉ}
{π ββ₯ π} {π β πΉ}
{π β₯ π}
Are on the graph of π(π) where the value (y-coordinate) is 0 or 1. EXCEPT π(π) is never 0! POINT is: (0, 1)
π = οΏ½π.πππ + π
3.
4. NR If the domain of the radical function π(π₯) = β23 β 5π₯ + 71 is π₯ β€ π, then the value of π, correct to the nearest tenth, is _______.
2.
Domain of οΏ½π(π) is defined by the zeros of π(π), as οΏ½π(π) is not defined between these points. (Not defined where π(π₯) is negative, canβt square root a negative!)
Invariant points where value of π(π) is 1 or 0. (4 total)
Domain of οΏ½π(π): π β€ βπ or π β₯ π
Range of οΏ½π(π): π β₯ π
Invariant points where π(π) is 1 or 0β¦. π(π) = π:
πππ β π = π
πππ = π
π = π
π(π) = π:
πππ β π = π
πππ = π
π = π *This question can also be solved graphically
23 β 5π₯ β₯ 0
23 β₯ 5π₯
235β₯ π₯ ππ: π₯ β€
23
5
Whatever is under the square root sign must be positive. (More specifically, βgreater than or equal to 0!β)
4.6
5. 6. MC: If π(π₯) = β3π₯ and π(π₯) = π₯2 + 2π₯ + 1, then an expression for π(π(π₯)) is:
A. 3π₯ + 2β3π₯ + 1
B. 9π₯2 + 2β3π₯ + 1
C. 3π₯ + β6π₯ + 1
D. 9π₯2 + β6π₯ + 1 7. If π(π₯) is a quadratic function in the form π¦ = ππ₯2 + ππ₯ + π with π > 0 and a vertex on the π₯-axis,
determine the domain and range of π¦ = οΏ½π(π₯).
Invariant points where value of π(π) is 1 or 0. (4 total)
One option is to graph the horizontal lines π¦ = 1 and π¦ = 0 and count the intersections!
= (β3π₯)2 + 2οΏ½β3π₯οΏ½ + 1
= (β3)2(π₯)2 + 2β3π₯ + 1
If the vertex is on the π₯-axis (and the lead coefficient π is positive) then π(π₯) is never negative. So the domain of
π¦ = οΏ½π(π₯) is all reals.
{π β πΉ}
{π β₯ π}
Like all equations in this course, radical equations can be solved either alnebraically, or nraphically. The solutions (or roots) of a radical equation are the same as the π₯-intercpets of the function.
1. Algebraically solve the following equations:
(a) οΏ½β5 β 3π₯οΏ½2
= (11)2 (b) 3 β 4β7 β 2π₯ = β13
β’ Alberta Ed Learning Outcome: Find the zeros of a radical function graphically and explain their β’ relationship to the π₯-intercepts of the graph and the roots of an equation.
Consider the equation βπ₯ β 4 = 3
We can solve this equation by:
Mathematical Reasoning:
Squarinn both sides:
First think β what number do we square root to get 3? Answer: 9
Then think, what number π₯ would we subtract 4 from to get 9?
Answer: π = ππ
Graphing:
οΏ½βπ β ποΏ½2
= (π)2
π₯ β 4 = 9
Our goal is to rid the left side of the square root sign, so that we can isolate π₯
Answer: π = ππ
Option 1
Graph π¦1 = βπ₯ β 4 & π¦2 = 3 Find point(s) of intersection
Option 2
Set equation to zero: βπ₯ β 4 β 3 = 0
Graph π¦1 = βπ₯ β 4 β 3 and find zeros.
Set π₯ max to some value greater than 10, since the solution must lay between the π₯ min and max.
π β ππ = πππ
βππ = πππ
π = βππππ
οΏ½5 β 3(β116
3) = 11
Check: substitute π₯ β 1163
back in the original equationβ¦
β121 = 11
First: isolate the square root term (Move the radical term to the Right Side so the lead
coefficient can be made positive.)
ππ = πβπ β ππ
(π)π = οΏ½βπ β πποΏ½π
ππ = π β ππ ππ = βπ
π = βππ
2. Use your graphing calculator to determine the π₯-intercept(s) of the functions. State any restrictions on the
variable. (a) π¦ = β1
2β2π₯ β 6 + 3 (b) π¦ = β2π₯2 + 1 β 11
3. Algebraically solve the following equations. Nearest hundredth where necessary.
(a) 12 β2π₯ β 6 = 3 (b) οΏ½β2π₯2 + 1οΏ½
2= (11)2
4. Solve the following equation algebraically and graphically.
(π₯ + 3)2 = οΏ½β2π₯2 β 7οΏ½2
You will have to adjust (enlarge) your window to see the π₯-intercept. Copy your graph and label the intercept here. (Provide a scale on each axis.)
π = ππ π β Β±π.ππ
Restriction: π β₯ π No Restriction
βππ β π = π
οΏ½βππ β ποΏ½π
= (π)π
ππ β π = ππ ππ = ππ
π = ππ
First: isolate the square root term (multiply both sides by β2β)
πππ + π = πππ
πππ = πππ
οΏ½ππ = βππ
π = Β±βππ
π β Β±π.ππ
ππ + ππ + π = πππ β π
π = ππ β ππ β ππ π = (π β π)(π + π)
π = π or βπ
( ) + 3 = οΏ½2( )2 β 7 π π
CHECK each solution:
11 = β121
(β ) + 3 = οΏ½2(π β )2 β 7 π
1 = β1
π¦1 = (π₯ + 3) β οΏ½2π₯2 β 7
π¦2 = 0
Graphically:
π = π