## Presentation on theme: "Chapter 2 Radical Functions."— Presentation transcript:

Square Roots In mathematics, a square root of a number a is a number y such that y2 = a For example, 4 is a square root of 16 because 42 = 16 And so is -4 because (-4) 2 = 16

So what is … = 2 … Why isn’t it 2 and -2? Because means the principal square root ... … the one that isn't negative! There are two square roots, but the radical symbol means just the principal square root

The square roots of 36 are …
… 6 and -6 But = … …6 When you solve the equation x2 = 36, you are trying to find all possible values that might have been squared to get 36

What about … −16 = ? There is no real number that when squared is negative For the purposes of Grade 12 Pre-Calculus you cannot take the square root of a negative number

Order??? Stretches and Reflections, performed in any order, followed by translations. If (𝑥,𝑦) is a point on 𝑓 𝑥 = 𝑥 then … … ( 1 𝑏 𝑥+ℎ,𝑎𝑦+𝑘) is a point on 𝑓 𝑥 =𝑎 𝑏(𝑥−ℎ) +𝑘

Graphing Radical Functions Using a Table of Values
A good place to start is to determine the domain The radicand must be greater than or equal to zero Remember to reverse the inequality when multiplying or dividing by a negative number

Graphing Radical Functions Using a Table of Values

The Domain is 𝑥 𝑥≤1,𝑥∈ℝ or −∞, 1
The Range is 𝑦 𝑦≥1,𝑦∈ℝ or 1, +∞

Not an invariant point!! It does not map to itself!!! (0,0) maps to (1,1) (1,1) maps to (0.5, 4)

Mapping Notation How points on this map to that

What do you notice? 𝑦= 𝑥 has been horizontally stretched by a factor of 𝑦= 𝑥 has been horizontally stretched by a factor of and vertically stretched by a factor of 2 𝑦= 𝑥 has been vertically stretched by a factor of 4 These all have the same graph!! They are identical!

So… “a” can do anything that “b” can do and vice versa … right? Can’t we just get rid of one of them? Wrong!!! Without “a” you can’t do reflections in the x-axis and without “b” you can’t do reflections in the y-axis.

Because the origin is an invariant point as far as stretching and reflecting is concerned we know that if the starting point hasn’t moved then no translations were involved. If the starting point of the graph hasn’t moved horizontally then “h” must be zero and if the starting point hasn’t shifted vertically then “k” must be zero. No amount of stretching or reflecting can change that. If the starting point has moved, then the values of h and k are just the coordinates of the place where the starting point has moved to.

Remember this!!

Since the graph has not been reflected in either the x or y axis we know that a and b must be positive.

It can be viewed as either a purely vertical stretch or purely horizontal stretch.

Viewed as a vertical stretch
Viewed as a horizontal stretch

Since the graph has been reflected in both the x or y axis we know that both a and b must be negative. So we can set a = -1 and find b or set b = -1 and find a.

The Domain is 𝑇 𝑇≥−273.15,𝑇∈ℝ or [ , +∞) The only transformations that can change the range as compared to the base function are vertical translations and reflections over the x-axis. Neither of these occur. The Range is 𝑠 𝑠≥0,𝑠∈ℝ or [0, +∞)

The graph of 𝑠= 𝑇 has been stretched vertically by a factor of about 20 and then translated horizontally about 273 units to the left.

Square Root of a function
First let’s look at …

You can use values of 𝑓(𝑥) to predict values of 𝑓 𝑥 and to sketch the graph of 𝑦= 𝑓 𝑥 .
The domain of 𝑦= 𝑓 𝑥 consists only of the values in the domain of 𝑓(𝑥) for which 𝑓(𝑥)≥0. The range of 𝑦= 𝑓 𝑥 consists of the square roots of all the values in the range of 𝑓(𝑥) for which 𝑓 𝑥 is defined. Invariant points occur at 𝑓 𝑥 =0 and 𝑓 𝑥 =1 because at these values 𝑓 𝑥 = 𝑓(𝑥) . For 𝑓 𝑥 >0 and 𝑓 𝑥 <1 the graph of 𝑓 𝑥 is above 𝑓 𝑥 and for 𝑓 𝑥 >1 the graph of 𝑓 𝑥 is below 𝑓 𝑥

The Domain of 𝑦= 𝑓(𝑥) is 𝑥 𝑥≤ 3 2 ,𝑥∈ℝ or −∞, 3 2
𝑦= 𝑓 𝑥 is only defined for 𝑓(𝑥)≥0. The Domain of 𝑦= 𝑓(𝑥) is 𝑥 𝑥≤ 3 2 ,𝑥∈ℝ or −∞, 3 2

Invariant points occur at 𝑓 𝑥 =0 and 𝑓 𝑥 =1 because at these values 𝑓 𝑥 = 𝑓(𝑥) .

x 𝒇 𝒙 =𝟑−𝟐𝒙 𝒚= 𝟑−𝟐𝒙 3 2 11 8 1 4 1 2 1 − 1 2 4 2 -3 9 3

The Domain of 𝑓(𝑥) is 𝑥 𝑥∈ℝ and the Range of 𝑓(𝑥) is 𝑦 𝑦∈ℝ
and the Range is 𝑦 𝑦≥0,𝑦∈ℝ

Invariant points occur at (1, 1) and (1.5, 0)

Find x-ints (set 𝑓 𝑥 =0): First find key points of 𝑓 𝑥 =2−0.5 𝑥 2 Vertex at (0, 2) and y-int = 2

Set 𝑓 𝑥 =1 Invariant points occur at 𝑓 𝑥 =0 and 𝑓 𝑥 =1 because at these values 𝑓 𝑥 = 𝑓(𝑥) . Invariant points occur at (-2, 0) , (2, 0) , (− 2 , 1), and ( 2 , 1)

The y-coordinates of the points on 𝑦= 𝑓(𝑥) are the square roots of the corresponding points on 𝑓(𝑥)

The Domain of 𝑓(𝑥) is 𝑥 𝑥∈ℝ and the Range of 𝑓(𝑥) is 𝑦 𝑦≤2, 𝑦∈ℝ
and the Range is 𝑦 0≤𝑦≤ 2 ,𝑦∈ℝ

The y-coordinates of the points on 𝑦= 𝑓(𝑥) are the square roots of the corresponding points on 𝑓(𝑥)

They are the same!!!!! Roots, zeroes, x-intercepts, solutions …??? What is the relationship between these things? The following phrases are equivalent: "find the zeroes of f(x)" "find the roots of f(x)" "find all the x-intercepts of the graph of f(x)" "find all the solutions to f(x)=0" Example: The roots of 𝑦= 𝑥+4 −1 are the zeroes of the function 𝑓(𝑥)= 𝑥+4 −1 are the solutions to the equation 𝑥+4 −1=0 and are the x-intercepts of the graph.

Determine the root(s) of 𝑥+5 −3=0 algebraically.
First consider any restrictions on the variable in the radical.

Determine the root(s) of 𝑥+5 −3=0 algebraically.
Algebraic solutions to radical equations sometimes produce extraneous roots In mathematics, an extraneous solution represents a solution that emerges from the process of solving the problem but is not a valid solution to the original problem. You must always check your solution in the original equation. Left Side Right Side

One of the basic principles of algebra is that one can perform the same mathematical operation to both sides of an equation without changing the equation's solutions. However, strictly speaking, this is not true, in that certain operations may introduce new solutions that were not present before. The process of squaring the sides of an equation creates a "derived" equation which may not be equivalent to the original radical equation.  Consequently, solving this new derived equation may create solutions that never previously existed.  These "extra" roots that are not true solutions of the original radical equation are called extraneous roots and are rejected as answers.

Solve the equation 𝑥+5 =𝑥+3 algebraically.

The solution is x = -1 Solve the equation 𝑥+5 =𝑥+3 algebraically.
Check: Left Side Right Side Left Side Right Side The solution is x = -1

Solve the equation 3 𝑥 2 −5 =x+4 graphically
Solve the equation 3 𝑥 2 −5 =x+4 graphically. Express your answer to the nearest tenth. First consider any restrictions on the variable in the radical.

Solve the equation 3 𝑥 2 −5 =x+4 graphically
Solve the equation 3 𝑥 2 −5 =x+4 graphically. Express your answer to the nearest tenth. Method 1: Graph each side of the equation as a function: 𝑓 𝑥 = 3 𝑥 2 −5 𝑔 𝑥 =𝑥+4 Then determine the values of x at the point(s) of intersection. The solutions are 𝑥≅−1.8 and 𝑥≅5.8

Solve the equation 3 𝑥 2 −5 =x+4 graphically
Solve the equation 3 𝑥 2 −5 =x+4 graphically. Express your answer to the nearest tenth. Method 2: Rearrange the radical equation so that one side is equal to zero: 3 𝑥 2 −5 −𝑥−4=0 Graph the corresponding function 𝑓 𝑥 = 3 𝑥 2 −5 −𝑥−4 And determine the x-intercepts of the graph. The solutions are 𝑥≅−1.8 and 𝑥≅5.8

Algebraic solutions sometimes produce extraneous roots, whereas graphical solutions do not produce extraneous roots. Algebraic solutions are generally exact while graphical solutions are often approximate.

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