Given the parabola y = x and y = (m-1) x + m, there are two intersections when m is a real number

Given the parabola y = x and y = (m-1) x + m, there are two intersections when m is a real number

Let x ^ 2 = (m ^ 2-1) x + m ^ 2
x^2-(m^2-1)x-m^2=0
(m^2-1)^2+4m^2>0
(m^2+1)>0
M is any real number
It's weird^^^^

Given that the intersection point of the parabola y = x2-x-1 and X axis is (m, 0), then the value of the algebraic formula m2-m + 2008 is______ ．

The intersection of y = x2-x-1 and x-axis is (m, 0), m2-m-1 = 0, that is m2-m = 1, the original formula = 1 + 2008 = 2009

Proof: no matter what real number m takes, the parabola y = x ^ + (m-5) x + M-8 and X-axis always have two intersections

It is proved that the discriminant of the function y = x ^ + (m-5) x + M-8 is △,
That is △ = B ^ 2-4ac
=(m-5)^2-4(m-8)
=m^2-10m+25-4m+32
=m^2-14m+57
=(m-7)^2+8>0
So no matter what real number m takes, there are always two intersections between the parabola y = x ^ + (m-5) x + M-8 and the X axis

Given that the origin is the highest point of the parabola y = (M + 1) X2, then the range of M is ()
A. m＜-1B. m＜1C. m＞-1D. m＞-2

∵ the origin is the highest point of the parabola y = (M + 1) X2, ∵ m + 1 ＜ 0, that is, m ＜ - 1

The point nearest to a (0, a) (a > 0) on the parabola y = 1 / 2 x ^ is the origin, and the value range of a is obtained,

Let the coordinates of a point P on the parabola be (x, y), then the distance L, L & sup2; = y & sup2; - 2ay + A & sup2; + 2Y from point P to point a is obtained by merging L & sup2; = y & sup2; - 2 (A-1) y + A & sup2; = [y - (A-1)] & sup2; + A & sup2; - (A-1) & sup2; let z = L & sup2

If the two intersections of the parabola y = - x square + (m-2) x + m + 1 and X axis are on the left side of the origin, then the value range of M is

M is less than - 1

The trajectory equation of the vertex of parabola y = x ^ 2 + 2mx + m ^ 2-2m (M belongs to R) is

It is known that the abscissa of the vertex of parabola is x = - B / 2a, and the ordinate is y = (4ac-b & sup2;) / 4A. Substituting a = 1, B = 2m, C = M & sup2; - 2m into x = - m y = [4 (M & sup2; - 2m) - (2m) & sup2;] / 4Y = (4m & sup2; - 8m-4m & sup2;) / 4Y = - 8m / 4Y = - 2m, so the vertex of parabola y = x ^ 2 + 2mx + m ^ 2-2m (M belongs to R)

It is known that the parabola y = - x square + MX-1, (M belongs to real number), when m changes, the trajectory equation of the focus of the parabola is?
The distance from vertex to focus should be 1 / 4

1, from the general formula of parabola vertex
X=-b/(2a)=-m/(-2)=m/2
Y=(4ac-b²)/4a=-(4-m²)/4
2. Eliminate m and get the equation of Y with respect to X
Y=1-X²
This is the trajectory of the vertex of the parabola
3, because the focus of the parabola is always 0.5 units below the vertex
So move the parabola vertex trajectory down 0.5 units
4. Replace y in 2 with y + 0.5
obtain
Y=-X²+0.5
(for mental arithmetic, LZ had better do it by himself:)
Yeah, yeah, yeah. It's 0.25
So the final equation is y = - X & sup2; + 0.75

Given the parabola C1: y = x ^ 2 + MX + 1 and C2: y = x ^ 2 + (1 / M) * x + 1, find the trajectory equation of the midpoint D of the vertex line of the two parabola

C1 vertex (- M / 2, (4-m ^ 2) / 4)
C2 vertex (- 1 / 2m, 1-1 / 4m ^ 2)
Point d (x, y)
2x=-m/2-1/2m
2y=(4-m^2)/4+1-1/4m^2
Eliminate M
y=-3/4-2x^2

Given the quadratic function y = x ^ 2-2x + 4, if there is only one intersection point between the line passing through the origin and the quadratic function, then there are several such lines

There should be three. Let the analytic expression of the straight line be y = KX. Then two solutions can be obtained simultaneously with the quadratic function. These are two, and the other one is x = 0