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Practice the technique with the following quadratics: (i) x? – 8x + 7 (ii) 2x2 – 5x + 1 (iii) Зx? — х — 4

Practice the technique with the following quadratics: (i) x? – 8x + 7 (ii) 2x2 – 5x + 1 (iii) Зx? — х — 4

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter3: Polynomial And Rational Functions
Section3.2: Polynomial Functinos And Their Graphs
Problem 2E: Describe the end behavior of each polynomial. (a) y=x38x2+2x15 End behavior: y as x y as x (b)...
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All quadratics can be expressed in the form a(x – h)² + k. For example, x? – 4x – 5 = (x – 2)² – 9 How to do it: a(x – h)² + k = a[(x – h)(x – h)] + k = a[x² – hx – hx + h²] + k = ax? – 2ahx + ah² + k So, we want to express x? – 4x – 5 in form a(x – h)² + k x² – 4x – 5 = a(x – h)² + k x2 – 4x – 5 = ax² – 2ahx + ah² + k Now equate the coefficients: 1 = a -4 = -2ah - 4 = -2(1)h Therefore, h = 2 -5 = h2 + k - 5 = 4 + k Therefore, k = -9 So x² – 4x – 5 = (x – 2)² – 9
Transcribed Image Text:All quadratics can be expressed in the form a(x – h)² + k. For example, x? – 4x – 5 = (x – 2)² – 9 How to do it: a(x – h)² + k = a[(x – h)(x – h)] + k = a[x² – hx – hx + h²] + k = ax? – 2ahx + ah² + k So, we want to express x? – 4x – 5 in form a(x – h)² + k x² – 4x – 5 = a(x – h)² + k x2 – 4x – 5 = ax² – 2ahx + ah² + k Now equate the coefficients: 1 = a -4 = -2ah - 4 = -2(1)h Therefore, h = 2 -5 = h2 + k - 5 = 4 + k Therefore, k = -9 So x² – 4x – 5 = (x – 2)² – 9
Practice the technique with the following quadratics: (i) х2 — 8х + 7 (ii) 2x2 — 5х + 1 (iii) 3x2 – x – 4
Transcribed Image Text:Practice the technique with the following quadratics: (i) х2 — 8х + 7 (ii) 2x2 — 5х + 1 (iii) 3x2 – x – 4
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