2 Square RootsIn mathematics, a square root of a number a is a number y such that y2 = aFor example, 4 is a square root of 16 because 42 = 16And so is -4 because (-4) 2 = 16
3 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
4 The square roots of 36 are … … 6 and -6But = ……6When you solve the equation x2 = 36, you are trying to find all possible values that might have been squared to get 36
5 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
6 Radical FunctionA radical function is a function that has a variable in the radicand
12 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 on𝑓 𝑥 =𝑎 𝑏(𝑥−ℎ) +𝑘
13 Graphing Radical Functions Using a Table of Values A good place to start is to determine the domainThe radicand must be greater than or equal to zeroRemember to reverse the inequality when multiplying or dividing by a negative number
14 Graphing Radical Functions Using a Table of Values
15 The Domain is 𝑥 𝑥≤1,𝑥∈ℝ or −∞, 1 The Range is 𝑦 𝑦≥1,𝑦∈ℝ or 1, +∞
16 Graphing Radical Functions Using Transformations
17 Not an invariant point!!It does not map to itself!!!(0,0) maps to (1,1)(1,1) maps to (0.5, 4)
19 What do you notice?𝑦= 𝑥 has been horizontally stretched by a factor of𝑦= 𝑥 has been horizontally stretched by a factor of and vertically stretched by a factor of 2𝑦= 𝑥 has been vertically stretched by a factor of 4These all have the same graph!! They are identical!
20 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.
22 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.
31 The Domain is 𝑇 𝑇≥−273.15,𝑇∈ℝor [ , +∞)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, +∞)
36 Square Root of a function First let’s look at …
37 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
38 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
39 Invariant points occur at 𝑓 𝑥 =0 and 𝑓 𝑥 =1 because at these values 𝑓 𝑥 = 𝑓(𝑥) .
51 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.
52 Determine the root(s) of 𝑥+5 −3=0 algebraically. First consider any restrictions on the variable in the radical.
53 Determine the root(s) of 𝑥+5 −3=0 algebraically. Algebraic solutions to radical equations sometimes produce extraneous rootsIn 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 SideRight Side
54 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.
56 The solution is x = -1 Solve the equation 𝑥+5 =𝑥+3 algebraically. Check:Left SideRight SideLeft SideRight SideThe solution is x = -1
57 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.
58 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 −5𝑔 𝑥 =𝑥+4Then determine the values of x at the point(s) of intersection.The solutions are 𝑥≅−1.8 and 𝑥≅5.8
59 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=0Graph the corresponding function𝑓 𝑥 = 3 𝑥 2 −5 −𝑥−4And determine the x-intercepts of the graph.The solutions are 𝑥≅−1.8 and 𝑥≅5.8
60 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.