If the parabola y = ax's square-6x passes through (2,0), what is the distance from the fixed point of the parabola to the origin of the coordinate?

If the parabola y = ax's square-6x passes through (2,0), what is the distance from the fixed point of the parabola to the origin of the coordinate?

The square of parabola y = ax - 6x passes through (2,0),
therefore
0=a*4-6*2
a=3
The parabola is as follows:
y=3x^2-6x
=3(x-1)^2-3
The vertex is (1, - 3)
Distance to origin = √ 1 + (- 3) ^ 2 = √ 10

It is known that the parabola y = x2 + BX + C passes through the origin and the distance between the two intersections of the parabola and the X axis is 3

Because the parabola goes through the origin, so C = 0
The distance between the intersection point of parabola and X axis is | x1-x2 | = ((x1 + x2) ^ 2-2x1x2) ^ (1 / 2) = | B | = 3
So the analytic formula of parabola is y = x ^ 2 + 3x or y = x ^ 2-3x

Given the origin of the parabola - 1, - 3, and the intersection of Y axis 0, - 5, find the analytic expression of the function

Because when x = 0, y = 0 or y = - 5
So let x = ay (y + 5)
Substituting (- 1, - 3) into
-1=a(-3)(-3+5)
a=1/6
So the analytic formula is x = 1 / 6y (y + 5)

The known set M = {- 1,1}, n = {1 / 2}
a

Because
a half

The known set M = {- 1,1} n = [Xi1 / 2]

From 1 / 2

If M = {x | x is less than 1} and N = {x | 2 is less than 1}, then m ∩ n is equal to

The power 0 of 2 = 1,2 > 1, so it is increasing, the power X of 2 is less than 1, so x is increasing

1. Set a = {x ∈ Z | X-1|

B is the answer
From |||||||||||||||||||||||||||||||
0 < X-1 ≤ 2
The solution is 1

Let set a = {x | y = (log2) (4-x ^ 2)}, B = {x | 4 / (x + 1)}. 1) find set a and CRB; 2) let inequality 2x ^ 2 + ax + B

Set a = {x | y = (log2) (4-x ^ 2)} = {x | 4-x ^ 2 > 0} = {x | 2}

The known set a = {x | log2 (6x + 12) > = log2 (x ^ 2 + 3x + 2), X ∈ r}, B = {x | 2 ^ (x ^ 2-m)

From log2 (6x + 12) > = log2 (x ^ 2 + 3x + 2), 6x + 12 > = x ^ 2 + 3x + 2 and x ^ 2 + 3x + 2 > 0 and 6x + 12 > 0 are obtained,
Get - 1 by solving inequality system

Let a = {X / x ^ 2 + 4x = 0}, B = {X / x ^ 2 + 2 (a + 1) x + A ^ 2-1 = 0}, a intersection B = B, and find the value of real number a
I see the solution of this problem on the Internet, a = plus or minus 1. But B is an empty set. Why not write it out?

The landlord is right
A={X|x^2+4x=0}={-4,0} ,
The intersection of a and B = B indicates that B is a subset of A
B is a quadratic equation
If there is no solution, it is an empty set
Discriminant = 4 [(a + 1) &# 178; - (A & # 178; - 1)]