The vertex form of f (x) = x^2-6x+5 is f (x) = a (x-h) ^2+k. What are the values of a, h, and k?

a = 1

h = 3

k = - 4

Step-by-step explanation:

To convert this standard form equation to vertex form, use complete the square.

f (x) = x² - 6x + 5

f (x) = (x² - 6x) + 5 <=group ax² and bx, factor out if needed, not in this case.

add and subtract (middle term/2) ²

f (x) = (x² - 6x + (6/2) ² - (6/2) ²) + 5 <=adding and subtracting the same number is like adding 0

f (x) = (x² - 6x + 9 - 9) + 5 <=simplify

f (x) = (x² - 6x + 9) - 9 + 5 <=take out the negative constant

f (x) = (x - 3) ² - 4 <=perfect square rule in brackets, simplify outside

a = 1 < = Nothing needed to be factored out from ax² and bx.

h = 3 <=If anyone says h = - 3, they are wrong. the negative is already in the general equation which says "x-h". It does not say "x+h".

k = - 4