Upload
marc-lambert
View
105
Download
0
Embed Size (px)
DESCRIPTION
Topics 2 and 3 from our class handout
Citation preview
𝒇(𝒙) =𝟏𝟐𝒙 + 𝟒
Alberta Ed Learning Outcome: Graph and analyze radical functions. • Transformations of radical functions also includes sketching and analyzing the transformation of
𝒚 = 𝒇(𝒙) to 𝒚 = �𝒇(𝒙). The function 𝒚 = 𝒇(𝒙) 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 (𝑥,𝑦) → (𝑥,�𝑦). Plot each new point. Sketch the resulting transformed graph.
A radical function can have the form 𝑦 = �𝑓(𝑥). In this topic we’ll examine the characteristics of the graph of a radical function, along with the domain and range.
𝒇(𝒙) = 𝒙𝟐
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 𝑦 = 𝑓(𝑥) and 𝑦 = �𝑓(𝑥)
Explain how you can derive the domain of a function
𝑦 = �𝑓(𝑥), given the graph of equaton of 𝑓(𝑥).
State the equation of the transformed function
State the domain and range of both 𝑦 = 𝑓(𝑥) and 𝑦 = �𝑓(𝑥)
Explain how the graph of 𝑦 = �𝑓(𝑥) differs from the graph of 𝑦 = 𝑥.
Invariant Points (where 𝒇 = 𝟎 or 𝟏)
(0,2) (10,3)
(0,4)
(10,16) 𝒚 = �𝟏𝟐𝒙 + 𝟒
𝒇(𝒙):
�𝒇(𝒙):
Domain: {𝒙 ∈ 𝑹}
Domain: {𝒙 ≥ −𝟖}
Range: {𝒚 ∈ 𝑹}
Range: {𝒚 ≥ 𝟎}
Excellent question! The function 𝒚 = �𝒇(𝒙) is only defined where 𝒇(𝒙) > 𝟎. So the domain can be found by locatnn the 𝒙-intercept of 𝒚 = 𝒇(𝒙) and determininn where the nraph is positiee (boie the 𝒙-axis)
𝒚 = �𝒙𝟐
(𝟒,𝟏𝟔)
(𝟒,𝟒)
All points (𝒙,𝒚) → (𝒙,�𝒚)
𝒚 = �𝟏𝟐𝒙 + 𝟒
𝒇(𝒙):
�𝒇(𝒙):
Domain: {𝒙 ∈ 𝑹}
Domain: {𝒙 ∈ 𝑹}
Range: {𝒚 ≥ 𝟎}
Range: {𝒚 ≥ 𝟎}
Very similar – howeier unlike on the nraph of 𝒇(𝒙), the nraph of �𝒇(𝒙) 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 𝑦 = 𝑓(𝑥), we
can obtain the graph of 𝑦 = �𝑓(𝑥) by transforming all
points (𝑥,𝑦) → (𝑥,�𝑦). See points ( and B
The domain of 𝑦 = �𝑓(𝑥) can be found by considering the zeros / 𝑥-intercepts of 𝑦 = 𝑓(𝑥).
Since we can’t square root negatives, 𝑦 = �𝑓(𝑥) is defined wherever 𝑓(𝑥) ≥ 0, that is, wherever the graph is above the 𝑥-axis. See point D
(D is the “start point” for the domain)
𝒇(𝒙) = −𝟐𝒙 + 𝟒
𝒚 = �𝒇(𝒙)
(−𝟐.𝟓,𝟗)
(−𝟐.𝟓,𝟏)
𝑨𝟏
𝑨𝟐 𝑩𝟏
𝑩𝟐 𝑪 𝑫
The invariant points in the transformation from 𝑦 = 𝑓(𝑥) to 𝑦 = �𝑓(𝑥) can be found by considering where the value of 𝑓(𝑥) 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 𝑦 = √4 − 2𝑥… The domain is:
And the invariant points occur wherever we are square rooting 0 or 1. (√0 = 0 and √1 = 1)
1st invariant point is where 𝑓(𝑥) = 0… 2nd invariant point is where 𝑓(𝑥) = 1…
Whatever we are square rooting cannot be negative. (That is, it must be ≥ 0)
4 − 2𝑥 ≥ 0
−2𝑥 ≥ −4
𝒙 ≤ 𝟐
-2 -2 *When dividing (or multiplying) both sides of an inequality by 0, reverse the inequality direction!
4 − 2𝑥 = 0
−2𝑥 = −4
𝒙 = 𝟐
4 − 2𝑥 = 1
−2𝑥 = −3
𝒙 = 𝟏.𝟓
Recall
𝑓(𝑥) is “4 − 2𝑥”
So, coordinates of invariant point are (𝟐,𝟎 ) So, coordinates of invariant point are (𝟏.𝟓,𝟏 )
1. For each given graph of 𝑦 = 𝑓(𝑥), sketch the graph of 𝑦 = �𝑓(𝑥), and state its domain, range, and any invariant points.
(a)
𝒇(𝒙) = 𝟐𝒙 − 𝟏
Domain of 𝑦 = 𝑓(𝑥): Domain of 𝑦 = �𝑓(𝑥):
Range of 𝑦 = 𝑓(𝑥): Range of 𝑦 = �𝑓(𝑥):
Invariant Points: 𝒚 = √𝟐𝒙 − 𝟏
(𝟔,𝟗)
(𝟔,𝟏)
{𝒙 ∈ 𝑹}
{𝒚 ∈ 𝑹}
{𝒙 ≥ 𝟏.𝟓}
{𝒚 ≥ 𝟎}
Are on the graph of 𝒇(𝒙) = 𝟐𝒙 − 𝟏 where the value (y-coordinate) is 0 or 1. POINTS are: and (1.5, 0) (2, 1)
(b)
𝒇(𝒙) = −𝟎.𝟓𝒙 + 𝟓
(c) 𝒇(𝒙) = 𝟎.𝟓𝒙𝟐 − 𝟐
Domain of 𝑦 = 𝑓(𝑥): Domain of 𝑦 = �𝑓(𝑥):
Range of 𝑦 = 𝑓(𝑥): Range of 𝑦 = �𝑓(𝑥):
Invariant Points:
Domain of 𝑦 = 𝑓(𝑥): Domain of 𝑦 = �𝑓(𝑥):
Range of 𝑦 = 𝑓(𝑥): Range of 𝑦 = �𝑓(𝑥):
Invariant Points:
(d) 𝒇(𝒙) = −𝒙𝟐 + 𝟏𝟔
Domain of 𝑦 = 𝑓(𝑥): Domain of 𝑦 = �𝑓(𝑥):
Range of 𝑦 = 𝑓(𝑥): Range of 𝑦 = �𝑓(𝑥):
Invariant Points:
(e) 𝒇(𝒙) = 𝟎.𝟓𝒙𝟐 + 𝟏
Domain of 𝑦 = 𝑓(𝑥): Domain of 𝑦 = �𝑓(𝑥):
Range of 𝑦 = 𝑓(𝑥): Range of 𝑦 = �𝑓(𝑥):
Invariant Points:
{𝒙 ∈ 𝑹}
{𝒚 ∈ 𝑹}
{𝒙 ≤ 𝟏𝟎}
{𝒚 ≥ 𝟎}
POINTS are: and (10, 0) (8, 1)
{𝒙 ∈ 𝑹}
{𝒚 ∈ 𝑹} {𝒙 ≤ −𝟐,𝒐𝒓 𝒙 ≥ 𝟐}
{𝒚 ≥ 𝟎}
Are on the graph of 𝒇(𝒙) where the value (y-coordinate) is 0 or 1. POINTS are: and (-2.45, 1), (-2, 0), (2, 0) (2.45, 1)
𝒚 = √−𝟎.𝟓𝒙 + 𝟓
On the graph of 𝒇(𝒙) the y-intercept is 5
…on 𝒚 = �𝒇(𝒙) it’s √𝟓, or approximately 2.24
𝒚 = �𝟎.𝟓𝒙𝟐 − 𝟐
DOM(IN: Graph of �𝑓(𝑥) is defined where 0.5𝑥2 − 2 ≥ 0
(Solve graphically – what are the x-intercepts of 𝑓(𝑥) = 0.5𝑥2 − 2 / where is the graph above the x-axis?)
0.5𝑥2 − 2 = 1 0.5𝑥2 = 3
𝑥 = ±�3
0.5
Find 𝑥 where 𝑓(𝑥) = 1
{𝒙 ∈ 𝑹}
{𝒚 ≤ 𝟏𝟔} {−𝟒 ≤ 𝒙 ≤ 𝟒}
{𝟒 ≤ 𝒚 ≤ 𝟎}
Are on the graph of 𝒇(𝒙) where the value (y-coordinate) is 0 or 1. POINTS are: and (-3.87, 1), (-4, 0), (4, 0) (3.87, 1)
−𝑥2 + 16 = 1 15 = 𝑥2
𝑥 = ±√15
Find 𝑥 where 𝑓(𝑥) = 1
𝒚 = �−𝒙𝟐 + 𝟏𝟔
{𝒙 ∈ 𝑹}
{𝒚 ∈≥ 𝟏} {𝒙 ∈ 𝑹}
{𝒚 ≥ 𝟏}
Are on the graph of 𝒇(𝒙) where the value (y-coordinate) is 0 or 1. EXCEPT 𝒇(𝒙) is never 0! POINT is: (0, 1)
𝒚 = �𝟎.𝟓𝒙𝟐 + 𝟏
3.
4. NR If the domain of the radical function 𝑓(𝑥) = √23 − 5𝑥 + 71 is 𝑥 ≤ 𝑘, then the value of 𝑘, correct to the nearest tenth, is _______.
2.
Domain of �𝒇(𝒙) is defined by the zeros of 𝒇(𝒙), as �𝒇(𝒙) is not defined between these points. (Not defined where 𝑓(𝑥) is negative, can’t square root a negative!)
Invariant points where value of 𝒇(𝒙) is 1 or 0. (4 total)
Domain of �𝒇(𝒙): 𝒙 ≤ −𝟏 or 𝒙 ≥ 𝟏
Range of �𝒇(𝒙): 𝒚 ≥ 𝟎
Invariant points where 𝒇(𝒙) is 1 or 0…. 𝒇(𝒙) = 𝟏:
𝟏𝟐𝒙 − 𝟏 = 𝟏
𝟏𝟐𝒙 = 𝟒
𝒙 = 𝟖
𝒇(𝒙) = 𝟎:
𝟏𝟐𝒙 − 𝟏 = 𝟎
𝟏𝟐𝒙 = 𝟏
𝒙 = 𝟔 *This question can also be solved graphically
23 − 5𝑥 ≥ 0
23 ≥ 5𝑥
235≥ 𝑥 𝑜𝑟: 𝑥 ≤
23
5
Whatever is under the square root sign must be positive. (More specifically, “greater than or equal to 0!”)
4.6
5. 6. MC: If 𝑓(𝑥) = √3𝑥 and 𝑔(𝑥) = 𝑥2 + 2𝑥 + 1, then an expression for 𝑔(𝑓(𝑥)) is:
A. 3𝑥 + 2√3𝑥 + 1
B. 9𝑥2 + 2√3𝑥 + 1
C. 3𝑥 + √6𝑥 + 1
D. 9𝑥2 + √6𝑥 + 1 7. If 𝑓(𝑥) is a quadratic function in the form 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 with 𝑎 > 0 and a vertex on the 𝑥-axis,
determine the domain and range of 𝑦 = �𝑓(𝑥).
Invariant points where value of 𝒇(𝒙) is 1 or 0. (4 total)
One option is to graph the horizontal lines 𝑦 = 1 and 𝑦 = 0 and count the intersections!
= (√3𝑥)2 + 2�√3𝑥� + 1
= (√3)2(𝑥)2 + 2√3𝑥 + 1
If the vertex is on the 𝑥-axis (and the lead coefficient 𝑎 is positive) then 𝑓(𝑥) is never negative. So the domain of
𝑦 = �𝑓(𝑥) is all reals.
{𝒙 ∈ 𝑹}
{𝒚 ≥ 𝟎}
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 𝑥-intercpets of the function.
1. Algebraically solve the following equations:
(a) �√5 − 3𝑥�2
= (11)2 (b) 3 − 4√7 − 2𝑥 = −13
• Alberta Ed Learning Outcome: Find the zeros of a radical function graphically and explain their • relationship to the 𝑥-intercepts of the graph and the roots of an equation.
Consider the equation √𝑥 − 4 = 3
We can solve this equation by:
Mathematical Reasoning:
Squarinn both sides:
First think – what number do we square root to get 3? Answer: 9
Then think, what number 𝑥 would we subtract 4 from to get 9?
Answer: 𝒙 = 𝟏𝟏
Graphing:
�√𝒙 − 𝟒�2
= (𝟏)2
𝑥 − 4 = 9
Our goal is to rid the left side of the square root sign, so that we can isolate 𝑥
Answer: 𝒙 = 𝟏𝟏
Option 1
Graph 𝑦1 = √𝑥 − 4 & 𝑦2 = 3 Find point(s) of intersection
Option 2
Set equation to zero: √𝑥 − 4 − 3 = 0
Graph 𝑦1 = √𝑥 − 4 − 3 and find zeros.
Set 𝑥 max to some value greater than 10, since the solution must lay between the 𝑥 min and max.
𝟓 − 𝟏𝒙 = 𝟏𝟐𝟏
−𝟏𝒙 = 𝟏𝟏𝟔
𝒙 = −𝟏𝟏𝟔𝟏
�5 − 3(−116
3) = 11
Check: substitute 𝑥 − 1163
back in the original equation…
√121 = 11
First: isolate the square root term (Move the radical term to the Right Side so the lead
coefficient can be made positive.)
𝟏𝟔 = 𝟒√𝟕 − 𝟐𝒙
(𝟒)𝟐 = �√𝟕 − 𝟐𝒙�𝟐
𝟏𝟔 = 𝟕 − 𝟐𝒙 𝟐𝒙 = −𝟗
𝒙 = −𝟗𝟐
2. Use your graphing calculator to determine the 𝑥-intercept(s) of the functions. State any restrictions on the
variable. (a) 𝑦 = −1
2√2𝑥 − 6 + 3 (b) 𝑦 = √2𝑥2 + 1 − 11
3. Algebraically solve the following equations. Nearest hundredth where necessary.
(a) 12 √2𝑥 − 6 = 3 (b) �√2𝑥2 + 1�
2= (11)2
4. Solve the following equation algebraically and graphically.
(𝑥 + 3)2 = �√2𝑥2 − 7�2
You will have to adjust (enlarge) your window to see the 𝑥-intercept. Copy your graph and label the intercept here. (Provide a scale on each axis.)
𝒙 = 𝟐𝟏 𝒙 ≈ ±𝟕.𝟕𝟓
Restriction: 𝒙 ≥ 𝟏 No Restriction
√𝟐𝒙 − 𝟔 = 𝟔
�√𝟐𝒙 − 𝟔�𝟐
= (𝟔)𝟐
𝟐𝒙 − 𝟔 = 𝟏𝟔 𝟐𝒙 = 𝟒𝟐
𝒙 = 𝟐𝟏
First: isolate the square root term (multiply both sides by “2”)
𝟐𝒙𝟐 + 𝟏 = 𝟏𝟐𝟏
𝟐𝒙𝟐 = 𝟏𝟐𝟎
�𝒙𝟐 = √𝟔𝟎
𝒙 = ±√𝟔𝟎
𝒙 ≈ ±𝟕.𝟕𝟓
𝒙𝟐 + 𝟔𝒙 + 𝟗 = 𝟐𝒙𝟐 − 𝟕
𝟎 = 𝒙𝟐 − 𝟔𝒙 − 𝟏𝟔 𝟎 = (𝒙 − 𝟖)(𝒙 + 𝟐)
𝒙 = 𝟖 or −𝟐
( ) + 3 = �2( )2 − 7 𝟖 𝟖
CHECK each solution:
11 = √121
(− ) + 3 = �2(𝟐 − )2 − 7 𝟐
1 = √1
𝑦1 = (𝑥 + 3) − �2𝑥2 − 7
𝑦2 = 0
Graphically:
𝒙 = 𝟖