Expain how you could write a quadratic function in factored form that would have a vertex X coordinate of 3 and two distinct roots?

Two distinct roots means two real solutions for x (the parabola needs to cross the x-axis twice)

Vertex form of a quadratic equation: (h,k) is vertex

y = a(x-h)^2 + k

The x of the vertex needs to equal 3

y = a(x-3)^2 + k

In order to have two distinct roots the parabola must be (+a) upward facing with vertex below the x-axis or (-a) downward facing with vertex above the x-axis. Parabolas are symmetrical so for an easy factorable equation make "a" 1 or -1 depending on if you want the upward/downward facing one.

y = (x-3)^2 - 1

Vertex (3,-1) upwards facing with two distinct roots 4 and 2

y = x^2 -6x + 9 - 1

y = x^2 -6x + 8

y = (x - 4)(x - 2)

Step-by-step explanation:

Please mark brainliest :) hope it helps

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

answer: you need the equation in the question so i can answer it

step-by-step explanation: