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A. Fiil in the boxes by following the steps found on the right side 1. f(x) = 3x² – 4x + 1 = O&² - Ox) +1 = 3(x² – *+D) +1– 3(0) Group together the terms containing x. Factor out a, the coefficient of x². Make the expression enclosed in parentheses a perfect square trinomial by completing SUBTRACT the constant to the constant the squares and term outside the parentheses in order to maintain equality to function. Note that the constants added the quadratic are on the same side of the equation. That is why we have to subtract, NOT add, the second constant. = 3(x² - x +;) +1– 3 (;) = 3(x – )² – O You should arrive at this expression. Express the perfect square trinomial as the square of the binomial and simplify. f(x) = 3 (x –) 16) = 3(x - }) - 2. y = x² – 4x – 10 = (x² – 4x) – 10 = (x² – 4x +D) y = (x – 2)² – O %3D - 10 - 0

A. Fiil in the boxes by following the steps found on the right side 1. f(x) = 3x² – 4x + 1 = O&² - Ox) +1 = 3(x² – *+D) +1– 3(0) Group together the terms containing x. Factor out a, the coefficient of x². Make the expression enclosed in parentheses a perfect square trinomial by completing SUBTRACT the constant to the constant the squares and term outside the parentheses in order to maintain equality to function. Note that the constants added the quadratic are on the same side of the equation. That is why we have to subtract, NOT add, the second constant. = 3(x² - x +;) +1– 3 (;) = 3(x – )² – O You should arrive at this expression. Express the perfect square trinomial as the square of the binomial and simplify. f(x) = 3 (x –) 16) = 3(x - }) - 2. y = x² – 4x – 10 = (x² – 4x) – 10 = (x² – 4x +D) y = (x – 2)² – O %3D - 10 - 0

Algebra for College Students
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter9: Polynomial And Rational Functions
Section9.4: Graphing Polynomial Functions
Problem 47PS
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Transforming Quadratic Functions A. Fiil in the boxes by folllowing the steps found on the right side 1. f(x) 3 Зх2 — 4х + 1 = [(x² - Dx) +1 = 3(x² – x+ D) +1 – 3(]) Group together the terms containing x. Factor out a, the coefficient of x. 4 Make the expression enclosed in %3D parentheses a perfect square trinomial by completing SUBTRACT the constant to the constant the squares and term outside the parentheses in order to maintain equality to function. Note that the constants added the quadratic are on the same side of the equation. That is why we have to subtract, NOT add, the second constant. = 3(2* -+) +1 – 3 () = 3(x – )² – O 4 You should arrive at this expression. Express the perfect square trinomial as the square of the binomial and simplify. F6) = 3 (x - })' -; 2. у %3D х? — 4х — 10 = (x2 – 4x) – 10 = (x2 – 4x +D) – 10 – 0 y = (x – 2)2 – O %3D %3D
Transcribed Image Text:Transforming Quadratic Functions A. Fiil in the boxes by folllowing the steps found on the right side 1. f(x) 3 Зх2 — 4х + 1 = [(x² - Dx) +1 = 3(x² – x+ D) +1 – 3(]) Group together the terms containing x. Factor out a, the coefficient of x. 4 Make the expression enclosed in %3D parentheses a perfect square trinomial by completing SUBTRACT the constant to the constant the squares and term outside the parentheses in order to maintain equality to function. Note that the constants added the quadratic are on the same side of the equation. That is why we have to subtract, NOT add, the second constant. = 3(2* -+) +1 – 3 () = 3(x – )² – O 4 You should arrive at this expression. Express the perfect square trinomial as the square of the binomial and simplify. F6) = 3 (x - })' -; 2. у %3D х? — 4х — 10 = (x2 – 4x) – 10 = (x2 – 4x +D) – 10 – 0 y = (x – 2)2 – O %3D %3D
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