Transformation properties #3 - Questions and Answers

Question 1 Given the parent function f(x) and descriptions of the transformations, write the equation of the transformed function, g(x). (Transformation should go in the described order.) f(x) = x³; vertical streetch by a factor 2.5, reflection over y-axis, horizontal translation leftl. Answer 1 Step 1 Given function is f(x) = x³ To write the equation of transformed function. Solution: Given function is f(x) = x³. We know that when y=f(x) is stretched vertically by c units then function becomes y=cf(x). So, when function f(x) = x³ is strectched vertically by 2.5 then function will becomes y = (2.5)x³ Step 1 Given function is f(x) = x³ To write the equation of transformed function. Solution: Given function is f(x) = x³. We know that when y=f(x) is stretched vertically by c units then function becomes y=cf(x). So, when function f(x) = x³ is strectched vertically by 2.5 then function will becomes y = (2.5)x³ Now, the graph of the function y=f(-x) is the graph of y=f(x) reflected in the y-axis. So, when y = (2.5)x³ isreflected in the y-axis then function will become y = (2.5)(-x)³ And if c is positive real number then graph of f(x-c) is the graph of y=f(x) shifted horizontally by c units. So, when y = (2.5) (-x)³ is shifted left by 1 unit then function will be g(x) = (2.5)(-x+1)³. Answer 2 To find the equation of the transformed function, we need to apply the given transformations to the parent function f(x) = x³ in the specified order. Let's go through each transformation step by step. Vertical stretch by a factor of 2.5: We multiply the parent function by 2.5. The equation becomes g(x) = 2.5.f(x). Reflection over the y-axis: We replace x with -x in the equation. The equation becomes g(x) = 2.5.f(-x). Horizontal translation left by 1 unit: We add 1 to x in the equation. The equation becomes g(x) = 2.5.f(-x + 1). Therefore, the equation of the transformed function g(x) is g(x) = 2.5.f(-x + 1) = 2.5.(-(x+1))³ Answer 3 Explanation: The parent function is given as f(x) = x³. We need to apply the following transformations to obtain the equation of the transformed function, g(x), in the described order: Vertical stretch by a factor of 2.5: To vertically stretch a function, we multiply the function by the stretch factor. Therefore, the first transformation can be written as g(x) = 2.5.f(x). Reflection over the y-axis: To reflect a function over the y-axis, we replace x with -x. Hence, the equation after the reflection becomes g(x) = 2.5.f(-x). Horizontal translation left 1 unit: To translate a function horizontally, we add or subtract a value from x. In this case, we translate left 1 unit, so we subtract 1 from x. Thus, the equation after the translation becomes g(x) = 2.5.f(-x-1). Combining all the transformations, the equation of the transformed function g(x) is: g(x) = 2.5.f(-x-1) = 2.5.(-(x+1))³ Answer 4 Step 1. Vertical stretch by a factor of 2.5: To vertically stretch the function, we multiply the parent function by the stretch factor. The equation becomes: g(x) = 2.5.f(x) Substituting the parent function f(x) = x³, we get: g(x) = 2.5.x³ Step 2. Reflection over the y-axis: To reflect a function over the y-axis, we replace x with -x in the equation. The equation becomes: g(x) = 2.5.(-x)³ Simplifying the exponent, we have: g(x) = 2.5.(-x)³ = 2.5(-xxx) = 2.5(-x³) Step 3. Horizontal translation left 1: To translate a function horizontally, we add the translation value to x. In this case, we're translating left 1 unit, so we add 1 to x. The equation becomes: g(x) = 2.5.(-(x + 1)³) Combining all the transformations, the final equation of the transformed function g(x) is: g(x) = 2.5.(-(x + 1)³) This equation represents a function obtained by vertically stretching the parent function f(x) = x³ by a factor of 2.5, reflecting it over the y-axis, and horizontally translating it left 1 unit. Question 2 Trying to go ABC is reflected across the X axis and then across the Y access which rotation is equivalent to this composition transformations transformations. Answer Step 1 let A(a,b) Step 2 if a point is reflected over x axis then the sign of ordinate change therfore A'(a,-b) and then point A' is reflected in y axis it cause the sign change of absissca therefore A"(-a,-b) the direct transformation of A to A" is reflection over origin. Question 3 The graph below shows transformations of the graph of f(x) = 2^x. What is the equation for the transformation? [Image of a graph of an exponential function, labeled y=f(x)] Answer Step 1 Given We have been given the graph of the equation which shows the transformation of f(x)=2^x. Step 2 To show We have to find the equation for the transformation. The general form of the equation of an exponential function is y = a(b)^x + c Here c represents the asymptote of the function. Step 3 Here the function has an asymptote y = 3 so, c = 3 and b = 2 as (2)^x. Hence we get y = a(2)^x + 3 From the graph, we have the coordinates (0, 1), and now substituting in the above equation we get 1 = a(2)⁰ + 3 1-3 = a(2)⁰ -2=a (as 2⁰ = 1) Hence we get the value of a = -2. Substituting the value in equation (1) we get y = (-2)(2)^x + 3 -3 1 2 0 2 -2 -1 Now we can see that the graph is reflected along the y-axis and for that, we will change x to -x and we get y = (-2)(2)^(-x) + 3 Now we get [Image of a graph of an exponential function, labeled y=f(x)] Question 4 A population with µ = 587 and σ = 12 is transformed into z-scores. After the transformation, what is the standard deviation for the population of z-scores? 2.A population with μ = 58 and σ = 12 is transformed into z-scores. After the transformation, what is the mean for the population of z-scores? Please help me answer these two! Answer 1 Step 1: z-score: The mean and standard deviation of the standard normal distribution is 0 and 1, respectively. The formula for z score is given as follows: z = (x - μ)/σ Step 2 1.Standard deviation for the population of z-scores: After the transformation, the standard deviation for the population of z-scores is 1. Answer 2 2.Mean for the population of z-scores: After the transformation, the mean for the population of z-scores is 0. Question 5 Prove that (a) the zero transformation and (b) the identity transformation are linear transformations. Answer a)Given: The linear transformation T: V → V represented as T(vᵢ) = 0 for i = 1, 2, ..., n. Approach: Consider an arbitrary v = v₁, v₂,..., vₙ is basis for V. The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties. The linear transformation is given by, T(vᵢ) = 0, i = 1,2,...,n... (1) The vector additinon is given by, T(u + v) = T(u) + T(v) The scalar multiplication is given by, T(cu) = cT(u) Calculation: As the vector set v is the subspace of V, the vector v can be written linear combination. Write the subspace vas linear combination. v = c₁v₁ + c₂v₂ + ... +cₙvₙ, (2) Here, c₁, c₂, ..., cₙ are arbitrary scalars. Conclusion: Hence, it is proved above that the set (v₁, v₂, ..., vₙ) is represented as v = c₁v₁ + c₂v₂ + ... + cₙvₙ b) Given: The linear transformation T : V → V represented as T(vᵢ) = 0 for i = 1, 2, ..., n. Approach: Consider an arbitrary v = v₁, v₂,..., vₙ is basis for V. The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties. The linear transformation is given by, T(vᵢ) = 0, i = 1,2,...,n... (1) The vector additinon is given by, T(u + v) = T(u) + T(v) The scalar multiplication is given by, T(cu) = cT(u) Calculation: As the vector set v is the subspace of V, the vector v can be written linear combination. Write the subspace vas linear combination. T(v) = T(c₁v₁ + c₂v₂ + ... + cₙvₙ) = T(c₁v₁) + T(c₂v₂) + ... + T(cₙvₙ) = c₁T(v₁) + c₂T(v₂) + ... +cₙT(vₙ) ... (3) Conclusion: The transformation form of linear combination v = c₁v₁ + c₂v₂ + ... + cₙvₙ is T(v) = c₁T(v₁) + c₂T(v₂) + ... + cₙT(vₙ) c) Given: The linear transformation T : V → V represented as T(vᵢ) = 0 for i = 1, 2, ..., n. Approach: Consider an arbitrary v = v₁, v₂,..., vₙ is basis for V. The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties. The linear transformation is given by, T(vᵢ) = 0, i = 1,2,...,n... (1) The vector additinon is given by, T(u + v) = T(u) + T(v) The scalar multiplication is given by, T(cu) = cT(u) The scalar multiplication is given by, T(cu) = cT(u) Calculation: Solve formula (3) with use of formula(1) T(v) = c₁T(v₁) + c₂T(v₂) + ... + cₙT(vₙ) = c₁(0) + c₂(0) + ... + cₙ(0) = 0 From above calculation is is clear linear transformation T:V → V satisfies T(vᵢ) = 0 for i