A triangle has vertices at A(-3, - 1), B(-6, - 5), C(-1, - 4). Which transformation would produce an image with vertices A'(3. – 1), B (6. – 5), C'(1. – 4)?

add an image

step-by-step explanation:

no, because the sides are not congruent

answer: your answer is a

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0