The analytical process of finding the parabola should be detailed and points should be given for cor

Solution: From y=mx 2+3mx-4m, it can be known:

When x=0, y=-4m is c(0,-4m); When y = 0, x = -4 or 1 i.e. a(-4,0) b(1,0) or a(1,0)b(-4,0).

Then the length of AB is 5 and the length of BC is also 5

1) When b(1,0)c(0,-4m) is obtained according to the two-point distance formula.

1+16m^2=25

The solution yields m = root 3 2 or root -3 2

2) When b(-4,0)c(0,-4m), the same can be obtained: 16+16m, 2=25m, no solution, so the coordinates of b cannot be (-4,0).

In summary: the analytic formula is y=root3 2x 2+3 times root3 2x-2 times root3 or y=-root3 2x 2-3 times root3 2x+2 times root3

Hope my landlord is satisfied! Have a great weekend!

y=mx²+3mx-4m=m(x²+3x-4)=m(x-1)(x+4)(m>0)

Let x=0, we get c(0,-4m).

Let y=0 to get a(-4,0), b(1,0), or b(-4,0), a(1,0).

When a(-4,0), b(1,0), 25=16m 2+1 is obtained from ab=bc, and m= (6 2) is obtained

Therefore y= (6 2)(x+4)(x-1);

When b(-4,0), a(1,0), 25=16m 2+16 is obtained by ab=bc

The solution is m=3 4

Therefore y=3 4(x+4)(x-1).

a（x1，0），b（x2，0）

x1+x2=-3，x1*x2=-4

AB square = (x1-x2) square = (x1 + x2) square - 4x1x2 = 9 + 16 = 25 = bc square = x2 square + (4m square).

Because c(0,-4m),-4m is less than 0

then a and b are on either side of the y-axis.

a（-2，0）b（3，0）x（0，-4）

That is, m=1y=x-square + 3x-4

Parabolic analytic method: according to the image to find the vertex coordinates (h, k) into the formula y=a(x-h) 2+k, and then find another point from the image to sit on the year of the rapid mark and substitute the above formula to find the quadratic function a. Or if you know the coordinates of any three points a, b, and c on the parabola, you can set the parabolic equation as y=ax +bx+c, and substitute the three points into the equation to solve the system of ternary equations to solve the values of a, b, and c, and finally obtain the parabolic equation.

Parabolic formula:

General formula: y=ax2+bx+c (a, b, c are constants, a≠ blind grip 0).

Vertex formula: y=a(x-h)2+k (a, h, k are constants, a≠0).

Intersection (two-pole): y=a(x-x1)(x-x2)(a≠0).

Parabolic vertex coordinates formula.

The vertex coordinates formula for y=ax +bx+c(a≠0) is (-b 2a, (4ac-b) 4a).

The vertex coordinates of y=ax +bx are (-b 2a, -b 4a).

Parabolic standard equations.

Right opening parabola: Y 2 = 2px

Left opening parabola: y 2 = -2px

Upper opening parabola: x 2 = 2py y = ax 2 (a is greater than or equal to 0).

Lower opening parabola: x 2 = -2py y = ax 2 (a less than or equal to 0).

p is the focal distance (p0)].

Peculiarity. In the parabola y 2 = 2px, the focus is (p 2,0) and the equation for the alignment is x = -p 2, eccentricity e = 1, range: x 0;

In the parabola y 2 = 2px, the focus is (p 2,0) and the equation for the alignment is x = -p 2, eccentricity e = 1, range: x 0;

In the parabola x 2=2py, the focal point is (0, p 2) and the equation for the alignment is y= -p 2, eccentricity e = 1, range: y 0;

In the parabola x 2=2py, the focal point is (0, p 2) and the equation for the alignment is y= -p 2, eccentricity e = 1, range: y 0;

Parabolic area arc length formula.

Area area=2ab 3

arc length abc

b^2+16a^2 )/2+b^2/8a ln((4a+√(b^2+16a^2 ))b)

Parabolic parametric equations.

The parametric equation for the parabola y 2=2px(p0) is:

x=2pt^2

y=2pt where the geometric meaning of the parameter p, the distance from the focal point of the parabola f(p 2,0) to the alignment x=-p 2 is called the focal parameter of the parabola.

General formula: y=ax2+bx+c (a, b, c are constants, a≠0).

Vertex formula: y=a(x-h)2+k (a, h, k are constants, a≠0).

Intersection (two-pole): y=a(x-x1)(x-x2)(a≠0).

where the parabola y=ax2+bx+c (a, b, c are constants, a≠0) and the coordinates of the intersection of the x-axis, that is, the two real roots of the equation ax2+bx+c=0.

Knowing that the intersection point of the parabola and the x-axis is (x1,0), (x2,0) can be set to the intersection point of the parabola.

y=a(x-x1)(x-x2)

Then solve a according to other conditions.

In this example, x1=-2, x2=1, and the only point of intersection is y=a(x+2)(x-1).

Because the parabola also substitutes its scattering through c(2,8).

Get: a*(2+2)*(2-1)=8

Therefore, the analytic formula for the parabola of a=2 is .

y=2(x+2)(x-1)

i.e. y=2x +2x-4

The method of the coefficient to be determined can be quickly imitated

General formula: y=ax 2+bx+c

The top mu spring fiber dot formula: y=a(x-k) 2+h

Zero-point forest (intersection formula): y=a(x-x1)(x-x2).

Knowing the three points, let y=ax 2+bx+c(a≠0), and substitute the three points to solve a, b, and c

Knowing the vertices (h,k) and another point, let y=a(x-h) 2+k(a≠0) and substitute the other point, solve a, and put parentheses.

Knowing the intersection point with the x-axis (m,0)(n,0), let y=a(x-m)(x-n)(a≠0), substitute another point, solve a, and put parentheses.

In mathematics, a parabola is a plane curve, it is mirror-symmetrical, and when the orientation is roughly U-shaped (it is still a parabola if it is in a different direction). It applies to any of several superficially different mathematical descriptions that can be proven to be exactly the same curve.

A description of a parabola involves a point (foci) and a line (quasi-line). The focus is not on the alignment. A parabola is the trajectory of a point in that plane that is equidistant from the alignment and focus.

Another description of a parabola is as a conic section, formed by the intersection of a conical surface and a plane parallel to the tapered busbar. The third description is algebra.

A line perpendicular to the alignment and passing through the focal point (i.e., a line that splits the parabola through the middle) is called the "axis of symmetry". The point on the parabola that intersects the axis of symmetry is called the "vertex" and is the point where the parabola bends most sharply. The distance between the vertex and the focal point, measured along the axis of symmetry, is the focal length.

"Straight lines" are parallel lines of a parabola and pass through the focal point.

Know 3 points or a vertex and a point.

The formula y=ax 2+bx+c

There are vertices. y=a（x-m）^2+c

Knowing the vertices, the parabolic equation can be set to y=a(x-m) +2m+2, and substituting a(-1,0) into 0=a(-1-m) +2(m+1) to obtain a=-2 (m+1).

So the analytic formula is y=[-2 (m+1)](x-m) +2m+2, I hope it can help you, I wish you progress in learning o( o

Let y=a(x-m) 2 +2m+2 vertex substitution point left mark (-1,0).

0=a(-1-m)^2 +2m+2

Solve a=(-2m-2) (m+1) 2

So y=((-2m-2) (m+1) 2)*(x-m) 2 +2m+2

1.The parabola y=-x +bx+c intersects with the x-axis ((Find the analytic formula for the parabola.

Known points can be obtained by substitution.

0=-1+b+c,0=-9-3b+c

It can be solved that b=-2 and c=3 are the original analytic equations.

y=-x²-2x+3

2.The vertices of the parabola are at (-1.).-2) and again (-2.).-1) To find the parabola, the analytic equation for the parabola is .

y+2)=2p(x+1), substitution point (-2,-1)-1+2=2p(-2+1), p=1 2

It's y+2=(x+1), and simplification is.

y=x²+2x-1

Since the axis of symmetry is y, it can be assumed that the analytic formula of the function is y ax 2+c, and because the points (0,4), (1,-2) are substituted into the equation, c=4, a=-6, the analytic formula of the function is y=-6x 2+4