f(x) = 5x2 + 2x
更新1:
f(x) = 5x^2 + 2x the five x is raised to the second power*
How do I solve this?
2016-06-21 2:05 pm
Find the y coordinate of the vertex, k, of the following equation.
f(x) = 5x2 + 2x
f(x) = 5x2 + 2x
回答 (2)
2016-06-21 2:10 pm
(h, k) is the vertex of the equation f(x) = a(x - h)² + k
The given function :
f(x) = 5x² + 2x
f(x) = 5[x² + (2/5)x]
f(x) = 5[x² + 2(1/5)x + (1/5)²] - 5(1/5)²
f(x) = 5[x + (1/5)]² - (1/5)
The vertex of the f(x) is (-1/5, -1/5).
The y-coordinate of the vertex, k = -1/5
The given function :
f(x) = 5x² + 2x
f(x) = 5[x² + (2/5)x]
f(x) = 5[x² + 2(1/5)x + (1/5)²] - 5(1/5)²
f(x) = 5[x + (1/5)]² - (1/5)
The vertex of the f(x) is (-1/5, -1/5).
The y-coordinate of the vertex, k = -1/5
2016-06-21 2:32 pm
5x^2+2x = 5(x^2 + 2/5 x)
x^2+2/5 x = (x+2/10)^2 - 4/100 = (x+1/5)^2 - 1/25
5(x^2 + 2/5 x) = 5(x+1/5)^2 - 5/25
= 5(x+1/5)^2 -1/5
Vertex is (-1/5, -1/5)
y coordinate of the vertex, k = -1/5
x^2+2/5 x = (x+2/10)^2 - 4/100 = (x+1/5)^2 - 1/25
5(x^2 + 2/5 x) = 5(x+1/5)^2 - 5/25
= 5(x+1/5)^2 -1/5
Vertex is (-1/5, -1/5)
y coordinate of the vertex, k = -1/5
收錄日期: 2021-04-18 15:09:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160621060556AAu42XJ