Math 12 - 2-Radical Functions

Math 12 - 2.1-Base Radical Function

a vertical stretch, if a<0 --> the graph of y=√x is reflected in the x-axis

b horizontal stretch, if b<0 --> the graph of y=√x is reflected in the y-axis

h x – translation, if h>0 --> the graph of y=√x is moved to the right h units.

if h<0 --> the graph of y=√x is moved to the left h units.

k y – translation, if k>0 --> the graph of y=√x is moved up k units.

if k<0 --> the graph of y=√x is moved down k units.

Graph and properties of the base of radical function y=√x- (x,y) intercept is (0,0) , it is also the left end point. No right end point

-Domain: {x : [ x≥0, x∈R }

-Range: {y : [ y≥0, y∈R}

Radical functions can be transformed by

a, b, h, k effects the base radicalfunction as in the next table.

y=√x→ y=a√b( x−h)+k

Exp. 2.1.1 Given y=√x→ y=2√4( x−3)+1 , explain how the radical base function transform to ? Then

plot the graphs?

Solution: Given y=√x→ y=2√4( x−3)+1

a = 2: Vertical stretch by the factor of 2

b = 4: Horizontal stretch by the factor of 4

h = 3: The graph of y=√x shifts 3 units to the right.

k = 1: The graph of y=√x shifts 1 unit up.

Plot the graphs of y=√ x , y=2√4(x−3)+1

Domain: of y=2√4( x−3)+1, {x : [ x≥3, x∈R}

Range :of y=2√4( x−3)+1, {y : [ y≥1, y∈R }

y=√ x

y=2√4 (x−3)+1

Exp. 2.1.2 Given y=√x→ y=−2√−3(x+2)−4 , explain how the radical base function transform to ?

Then plot the graphs?

Solution: Given y=√x→ y=−2√−3(x+2)−4

a = -2: Vertical stretch by the factor of 2, reflected in the x-axis

b = -4: Horizontal stretch by the factor of 4, reflected in the y-axis

h = -2: The graph of y=√x shifts 2 units to the left.

k = -4: The graph of y=√ x shifts 4 unit down.

Plot the graphs of y=√ x , y=−2√−3(x+2)−4

Domain : of y=2√4( x−3)+1, {x : [ x≤−2, x∈R}

Range :of y=2√4( x−3)+1, {y : [ y≤−4, y∈R }

y=−2√−3( x+2)−4

y=√ x

Exp. 2.1.3 Given the transformed radical function y=−2√−(x−3) , plot the graph and explain how it is

transformed ? Determine the domain and range?

Solution: y=a√b(x−h)+k

a = -2: The base radical function is vertically stretched by the factor of 2, and is reflected in the x-axis.

b = -1: It is reflected in the y-axis.

h = 3: Then it shifted 3 units to the right.

The domain is {x : [ x≤3, x∈R } , the range is {y : [ y≥0, y∈R}

Plot the graph as in the following figure.

y=−2√−( x−3)

Exp. 2.1.4 Given the transformed radical function y=2√3( x+2) , plot the graph and explain how it is

transformed ? Determine the domain and range?

Solution: y=a√b(x−h)+k

a = 2: The base radical function is vertically stretched by the factor of 2.

b = 3: It is horizontally stretched by the factor of 3.

h = -2: Then it shifted 2 units to the left.

The domain is {x : [x≥−2, x∈R } , the range is {y : [ y≥0, y∈R}

Plot the graph as in the following figure.

y=2√3( x+2)

Exp. 2.1.5 The following transformations are applied to the base radical function y=√( x)

Determine the resulting radical function? Plot it's graph? Determine the domain and range?

Vertical stretch Horizontal stretch Reflection in Translation

By a factor of 2 By a factor of 1/3 the x-axis and the y-axis 2 units left and 5 units down

Solution: y=a√b(x−h)+k

a = -2: The base radical function is vertically stretched by the factor of 2, reflected in the x-axis

b = -1/3: It is horizontally stretched by the factor of 1/3, reflected in the y-axis

h = -2: Then it shifted 2 units to the left.

k = -5: Then it shifted 5 units down.

So y=a√b(x−h)+k → y=−2√−13

( x+2)−5

The domain is {x : [x≤−2, x∈R } , the range is {y : [ y≤−5, y∈R }

Plot the graph as in the following figure.

y=−2√−13

( x+2)−5

Exp. 2.1.6 The following transformations are applied to the base radical function y=√( x)

Determine the resulting radical function? Plot it's graph? Determine the domain and range?

Vertical stretch Horizontal stretch Reflection in Translation

By a factor of 4 By a factor of 1/2 the x-axis and the y-axis 6 units left and 5 units up

Solution: y=a√b(x−h)+k

a = -4: The base radical function is vertically stretched by the factor of 4, reflected in the x-axis

b = -1/2: It is horizontally stretched by the factor of 1/2, reflected in the y-axis

h = -6: Then it shifted 6 units to the left.

k = 5: Then it shifted 5 units up.

So y=a√b(x−h)+k → y=−4√−12

( x+6)+5

The domain is {x : [ x<−6, x∈R} , the range is {y : [ y<5, y∈R }

Plot the graph as in the following figure.

y=−4√−12

( x+6)+5

Exp. 2.1.7 Match the graphs and the radical functions ?

Solution:

Graph - A =

Graph - B =

Graph – C =

Graph – D =

Radical Function B : y=3√−3( x−3)+3

Radical FunctionC : y=−3√3(x−3)+3

Radical Function D : y=−3√−3(x−3)+3

Radical Function A : y=3√3( x−3)+3

Radical Function B : y=3√−3( x−3)+3

Radical FunctionC : y=−3√3(x−3)+3

Radical Function A : y=3√3( x−3)+3

Radical Function D : y=−3√−3(x−3)+3

Graph A Graph B

Graph C Graph D

Exp. 2.1.8 Plot the graph of the given radical function y=−3√(x+2)−3 , determine the domain and

range?

Solution: Plot the graph

The domain is {x : [x≥−2, x∈R } , the range is {y : [ y≤−3, y∈R }

Exp. 2.1.9 A graph is given, determine the radical function in term of y=a√b(x−h)+k , determine the

domain and range?

y=−3√( x+2)−3

Solution: Step1: From the left end point(5,-4), y=a√b(x−h)+k → y=a√b( x−5)−4

Step2: View the graph as vertical stretch, take the point (6,-8) to find a,

y=a√(x−5)−4→−8=a√(6−5)−4→a=−4

Step3: Determine the radical function equation: --> y=−4√( x−5)−4

Step4: Determine the domain and range:

The domain is {x : [ x≥5, x∈R }

The range is {y : [ y≤−4, y∈R}

Exp. 2.1.10 A graph is given, determine the radical function equation in term of y=a√b(x−h)+k ,

determine the domain and range?

Solution: Step1: From the right end point(2,1), y=a√b(x−h)+k → y=a√b( x−2)+1

Step2: It makes more sense to view the graph as horizontal stretch, instead of as vertical stretch due the term

√b( x−2) ,( x−2)<0where x≤2 , we need negative b to make square root real.

Take the point (-2,-8) to find b --> y=√b(x−2 )+1→−8=√b(−2−2)+1→b=−814

Step3: Determine the radical function equation: --> y=√−814

( x−2)+1

Plot the graph to confirm if y=√−814

( x−2)+1 is the equation of the given graph as shown in the

following figure.

y=√−814

(x−2)+1

We need reflection in the x-axis to get the exact graph to be the same as the given graph, so

y=√−814

(x−2)+1→ y=−√−814

( x−2)+1

Plot the graph of y=−√−814

( x−2)+1 one more time to final confirm if it is the radical function

equation for the given graph as shown in the next figure.

Step4: Determine the domain and range:

The domain is {x : [ x≤2, x∈R }

The range is {y : [ y≤1, y∈R }

y=−√−814

( x−2)+1

Math 12 - 2.2-Plotting graphs y=√x and y= f ( x)

Exp. 2.2.1 Compare the graph of y=2x+1 and y=√(2x+1) , determine the domain and range

of each equation, determine invariant points?

Solution: Plot the graphs of y=2x+1 and y=√(2x+1)

Domain Range

y=2x+1 {x : [ x∈R } {y : [ y∈R}

y=√(2x+1) {x : [ x≥−0.5, x∈R} {y : [ y≥0, y∈R}

Invariant points = The points those are the same

for y = f(x) or the points those are on the line y = x

so that point is (2.375, 2.375)

y=√(2x+1)

Exp. 2.2.2 Given y=x2+2x+3 and y=√( x2+2x+3) , determine the domain and range of

each equation ?

Solution: Plot the graphs of y=x2+2x+3 and y=√( x2+2x+3)

y=x2+2x+3→ y=x2+2x+1−1+3=( x+1)2+2 ,

y=√( x2+2x+3)→ y=√( x2+2x+1−1+3)=√(x+1)2+2

y=√(2x+1)

y=x2+2x+3

y=√( x2+2x+3)

Domain Range

y=x2+2x+3 {x : [ x∈R } {y : [ y≥2, y∈R }

y=√( x2+2x+3) {x : [ x∈R } {y : [ y≥1.375, y∈R}

Exp. 2.2.3 Given a graph as shown in the figure, determine the equation and plot the graph of its square root function? Find the domain and range of its

square root function?

Solution: y =ax + b

a=( y2−y1)( x2−x1)

=(−1−0)(0−(−1))

=−1→ y=−x+b

At (-1,0) --> 0=−(−1)+b→b=−1 , so y= f (x)=−x−1 , g (x)=√ f ( x)=√(−x−1)

Plot the graph of g (x)=√ f (x)=√(−x−1)

y= f ( x)=−x−1

g (x)=√ f ( x)=√(−x−1)

Domain Range

g (x)=−x−1 {x : [ x∈R } {y : [ y∈R}

g (x)=√(−x−1) {x : [ x≤−1, x∈R } {y : [ y≥0, y∈R}

Exp. 2.2.4 Given a graph as shown in the figure, determine the equation and plot the graph of its square root function? Find the domain and range of its square root function?

Solution: The bottom point of the graph = (-2,3), it's a parabolic function

y=ax2→ y=a( x+2)2+3

at (0,7) --> y=a( x+2)2+3→7=a(0+2)2+3→a=1 , so

y=a( x+2)2+3→ y= f ( x)=( x+2)2+3

g ( x)=√ f (x)→ g ( x)=√(x+2)2+3

Plot the graph of g ( x)=√(x+2)2+3 as shown in the figure.

Domain Range

y=(x+2)2+3 {x : [ x∈R } {y : [ y≥3, y∈R}

g (x)=√(x+2)2+3 {x : [ x∈R } {y : [ y≥1.75, y∈R}

Exp. 2.2.5 Given a graph as shown in the figure, determine the equation and plot the graph of its square root function? Find the domain and range of its square root function?

g ( x)=√(x+2)2+3y=( x+2)2+3

Solution: The top point of the graph = (-2,1),

it's a parabolic function

y=ax2 → y=a(x+2)2+1

at (-1,0) --> y=a(x+2)2+1→ 0=a(−1+2)2+1→a=−1 ,

so y=a(x+2)2+1→ y= f ( x)=−(x+2)2+1

g (x)=√ f ( x)→ g( x)=√−( x+2)2+1

Plot the graph of g (x)=√−( x+2)2+1 as shown in the figure.

Domain Range

y=−( x+2)2+1 {x : [ x∈R } {y : [ y≤1, y∈R }

g (x)=√−( x+2)2+1 {x : [−3≤x≤−1 } {y : [ 0≤y≤1}

g ( x)=√−( x+2)2+1

y= f ( x)=−( x+2)2+1

Exp. 2.2.6 Given a graph as shown in the figure, determine the equation and plot the graph of its square root function? Find the domain and range of its square root function?

Solution: y =ax + b

a=( y2−y1)( x2−x1)

=(0−(−2))(2−0)

=1→ y=x+b

At (2,0) --> 0=(2)+b→b=−2 ,

so y= f ( x)=x−2 , g (x)=√ f ( x)=√(x−2)

Plot the graph of g ( x)=√ f (x)=√( x−2)

Domain Range

y=x−2 {x : [ x∈R } {y : [ y∈R}

g (x)=√(x−2) {x : [ x≥2 } {y : [ y≥0}

y= f ( x)=x−2

g (x)=√ f ( x)=√(x−2)

Exp. 2.2.7 Prove that f ( x)=√(x+3) is not always less than g (x)=( x+3) ?

Solution: Plot both graphs of f ( x)=√(x+3) and g ( x)=( x+3) as shown in the figure.

It's quite clear that f ( x)=√(x+3) > g (x)=( x+3) where −3<x<−2

g (x)=( x+3)f ( x)=√(x+3)

Math 12 - 2.3-Graphically Solving Radical Equations

Radical Function Equations

Algebraicallysolving

Verify the

restrictions

Separatethe

radical

Square both sidesAnd

Solve for x

Last checkwith the

restrictions

Radical Function Equations Graphically Solving

Single Equation

Graph

Find the zeros of the function

Two Equations

Graph

Graph each Side, find

intersection points

y=√( x+3)−x−3

1+√(x+3)=x+41+√(x+3)=x+4

y=1+√(x+3)

y=x+4

Exp. 2.3.1 Given 3+√(x−1)=x

(a) Find x = ? by algebraically solving. (b) Find x = ? by graphically solving.

Solution: 3+√(x−1)=x

(a) By algebraically solving

Step1: Verify restriction: x≥1

Step2: Separate the radical: √( x−1)=x−3

Step3: Square both sides: √( x−1)2=(x−3)2

Step4: Solve for x: √( x−1)2=(x−3)2 → x−1=x 2−6x+9

x2−7x+10=(x−5)(x−2)=0→ x=2,5

(b) By graphically solving

- By a single equation graph. -By two equations graphs.

Plot the graph of y=√( x−1)+3−x Plot y=√( x−1) and y=x−3

The zero of the function = (5 , 0) The intersection point = (5 , 0)

So x = 5

y=√( x−1)

y=x−3

y=√( x−1)+3−x

Exp. 2.3.2 By algebraically solving, find x from y=√( x−2)+3 ?

Solution: y=√( x−2)+3

By algebraically solving, y = 0, so y=√( x−2)+3→0=√(x−2)+3

√( x−2)=−3→[√( x−2)]2=(−3)2 → x−2=9→ x=11

Exp. 2.3.3 By algebraically solving, find x from 3√( x+7)+6=15 ?

Solution: 3√( x+7)+6=15

By algebraically solving, 3√( x+7)+6=15→√(x+7)=3

√( x+7)=3→[√( x+7)]2=(3)2 → x+7=9→ x=2

Exp. 2.3.4 By graphically solving, find x from y=√( x−3)+2 ?

Solution: Plot the graph of y=√( x−3)+2 x = 7

Exp. 2.3.5 By graphically solving, find x from y=√( x+7)−3 ?

Solution: Plot the graph of y=√( x+7)−3 , x = 2

Exp. 2.3.6 By graphically solving, find x from y=−√( x+3)−1 ?

Solution: Plot the graph of , y=−√( x+3)−1 x = No solution for this equation.

Exp. 2.3.7 Given √( x−1)=3

(a) Find x = ? (b) What is the restriction of this equation? (c) Plot the graph?

Solution: √( x−1)=3 plot the graphs of y=√( x−1) and y=3

(a) x = 10

(b) The restriction is x≥1

(c) Plot the graph as the figure.

Exp. 2.3.8 Given 2√(1+3x)=8

(a) Find x = ? (b) What is the restriction of this equation? (c) Plot the graph?

Solution: 2√(1+3x)=8→√(1+3x)=4 ,plot the graphs of y=√(1+3x) and y=4

(a) x = 5

(b) The restriction is x≥−0.67

(c) Plot the graph as the figure.

Exp. 2.3.9 Given 3+√(−4x+12)=x

(a) Find x = ? by algebraically solving (b) What is the restriction of this equation?

Solution: (a) 3+√(−4x+12)=x→√(−4x+12)=x−3

√(−4x+12)=x−3→[√(−4x+12)]2=(x−3)2 →−4x+12= x2−6x+9

x2−2x−3=0→( x+1)( x−3)=0→ x=−1,3

If x = -1, 3+√(−4x+12)=x→ 3+√(−4(−1)+12)=−1→ 3±4=−1

If x = 3, 3+√(−4x+12)=x→ 3+√(−4(3)+12)=3→3=3

So x = 3

(b) The restriction is x≤3

Exp. 2.3.10 Given √(5x+11)−3=x

(a) Find x = ? by algebraically solving (b) What is the restriction of this equation?

Solution: (a) √(5x+11)−3=x→√(5x+11)=x+3→[√(5x+11)]2=( x+3)2

[√(5x+11)]2=(x+3)2 →5x+11=x2+6x+9→ x2+x−2=0=( x−1)( x+2)

x=1,−2

If x = 1, √(5x+11)−3=x→√(5(1)+11)−3=1→1=1

If x = -2, √(5x+11)−3=x→ √(5(−2)+11)−3=−2→−2=−2

So x = 1, -2

(b) The restriction: √(5x+11)−3=x→5x+11≥0→ x≥−2.2

Exp. 2.3.11 Given 4=√(−x+2)+6 , find x = ? Is there a restriction of this equation?

Solution: 4=√(−x+2)+6→ √(−x+2)=−2

√(−x+2)=−2→[√(−x+2)]2=(−2)2 →−x+2=4→ x=−2

Replace x in the original equation: 4=√(−x+2)+6→ 4=√(−(−2)+2)+6→ 4=±2+6

Plot the graphs of y=√(−x+2) and y=−2

There is no intersection point between the 2 graphs. No solution.

Exp. 2.3.12 Given f ( x)=−√( x)−2 , find x = ? by algebraically solving and confirm by graphically

solving. Show the restrictions if there are any of them?

Solution: f ( x)=−√( x)−2→−√( x)−2=0→[−√( x)]2=(2)2 → x=4

Plot the graph of f ( x)=−√( x)−2

The graph shows NO SOLUTION.