Please help me to transform this to vertex form: y=5x^2-2x+2?
回答 (3)
2016-06-22 8:43 pm
y = 5x² - 2x + 2
y = 5[x² - (2/5)x] + 2
y = 5[x² - 2(1/5)x + (1/5)²] - 5(1/5)² + 2
The vertex form :
y = 5[x - (1/5)]² + (9/5)
y = 5[x² - (2/5)x] + 2
y = 5[x² - 2(1/5)x + (1/5)²] - 5(1/5)² + 2
The vertex form :
y = 5[x - (1/5)]² + (9/5)
2016-06-22 8:45 pm
y = 5x² - 2x + 2
factor out the leading coefficient
y = 5(x² - ⅖x) + 2
complete the square
coefficient of the x term: -⅖
divide it in half: -⅕
square it: (-⅕)² = (⅕)²
use (⅕)² to complete the square:
y = 5(x² - ⅖x + (⅕)²) - 5(⅕)² + 2
= 5(x - ⅕)² - ⅕ + 2
= 5(x - ⅕)² + 1⅘
vertex: (⅕, 1⅘)
factor out the leading coefficient
y = 5(x² - ⅖x) + 2
complete the square
coefficient of the x term: -⅖
divide it in half: -⅕
square it: (-⅕)² = (⅕)²
use (⅕)² to complete the square:
y = 5(x² - ⅖x + (⅕)²) - 5(⅕)² + 2
= 5(x - ⅕)² - ⅕ + 2
= 5(x - ⅕)² + 1⅘
vertex: (⅕, 1⅘)
2016-06-22 9:54 pm
y = 5 [ x² - (2/5) x ] + 2
y = 5 [ x² - (2/5) x + 1/25 ] - 1/5 + 2
y = 5 [ x - 1/5 ] ² + 9/5
y = 5 [ x² - (2/5) x + 1/25 ] - 1/5 + 2
y = 5 [ x - 1/5 ] ² + 9/5
收錄日期: 2021-04-18 15:13:45
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https://hk.answers.yahoo.com/question/index?qid=20160622124052AA8cSSm