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

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles