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