Quadratic function and quadratic equation of one variable If there are two intersections between quadratic function y = mx2-2x-1 and X axis, what is the value range of M? m>-1 ,m≠0.

Quadratic function and quadratic equation of one variable If there are two intersections between quadratic function y = mx2-2x-1 and X axis, what is the value range of M? m>-1 ,m≠0.

First of all, you need to understand that quadratic function and bivariate equation are actually corresponding. They can be transformed into each other. Quadratic function y = mx2-2x-1 can be regarded as bivariate equation mx2-2x-1 (only this bivariate equation has a specific value, that is y). So: because quadratic function y = mx2-2x-1 and X axis have two
The equations are as follows
y=mx2-2x-1
Y=0
When y = 0 (x-axis) is taken into y = mx2-2x-1, we get
mx^2-2x-1=0
This is a quadratic equation of one variable about X. its solution is the intersection of quadratic function and X axis,
There should be two unequal real roots, so the discriminant must be greater than zero
Δ=(-2)^2+4m=4m+4>0,m>-1
In addition, when m = 0, the quadratic function becomes a linear function and the graph becomes a direct expansion
The equations are as follows
y=mx2-2x-1
Y=0
When y = 0 (x-axis) is taken into y = mx2-2x-1, we get
mx^2-2x-1=0
This is a quadratic equation of one variable about X. its solution is the intersection of quadratic function and X axis,
There should be two unequal real roots, so the discriminant must be greater than zero
Δ=(-2)^2+4m=4m+4>0,m>-1
In addition, when m = 0, the quadratic function becomes a first-order function, and the graph becomes a straight line, and the straight line and the straight line can not have two intersections (either only one intersection, or no intersection, or infinitely many intersections (the same line)), so the value range of M is: M > - 1, m ≠ 0
On the problem of mathematical quadratic function equation of one variable
In the quadratic function y = - x + (2m + 2) x - (M's quadratic + 4m-3) with X as the independent variable, M is an integer not less than 0. Its image intersects with X axis at points a and B. point a is to the left of the origin and point B is to the right of the origin
1. Find the analytic solution of this quadratic function
2. The image of the first-order function y = KX + B passes through point a, intersects with the image of the second-order function at point C, and the area of triangle ABC is 10
1. M is obtained from the discriminant > 0
It is known that the line L: 3x-4y-1 = 0 intersects the original C: x ^ 2 + y ^ 2-2x + 4y-4 = 0 at two points a and B, and the area of triangle ABC is calculated
origin
Circle C: x ^ 2 + y ^ 2-2x + 4y-4 = 0
（x-1）^2+(y+2)^2=9
Center (1, - 2) radius = 3
Distance from center of circle to straight line = | 3 + 8-1 | / 5 = 2
|AB|=2√5
Distance from origin to line = 1 / 5
Area of triangle ABC = (1 / 2) * 1 * | ab | = √ 5
What point is C?
The distance from the origin to the line is the height of the triangle
The length of AB can be obtained
The formula of encounter problem
The basic quantitative formula of travel problem is: speed × time = distance △ speed = time distance △ time = speed 1. Encounter problem: sum of speed × encounter time = distance between two places △ sum of speed = distance between two places
The straight line: 3x-3y-5 = 0 intersects the circle C: (X-2) ^ 2 + (Y-1) ^ 2 = 25 at two points a and B. calculate the area of △ ABC
It should be 3x-4y
Using point difference method~
It's very easy to form a system of equations from the equations of a line and a circle, and find out a set of solutions to sweep the coordinates of points a and B. A. The coordinates of the three points B and C are all available, and it is easy to obtain the area of the besieged city by using the space analytic geometry. You can also calculate the distance between ab after calculating the AB coordinate. This is the bottom of the triangle. When calculating the distance between C and the straight line, this is the height of the triangle. When calculating the area, it's easy to expand
It's very easy to form a system of equations from the equations of a line and a circle, and find out a set of solutions to sweep the coordinates of points a and B. A. The coordinates of the three points B and C are all available, and it is easy to obtain the area of the besieged city by using the space analytic geometry. Also, after calculating the AB coordinate, calculate the distance between AB, which is the bottom of the triangle. After calculating the distance between C and the line, which is the height of the triangle, it is easy to calculate the area
Center C (2,1), radius r = 5
The distance h between C and the straight line is the height on the bottom ab of △ ABC: H = | 3 * 2 - 4 * 1 - 5 | / √ (3 ^ 2 + 4 ^ 2) = 3 / 5
|AB| = 2√(r^2 -h^2) = 2√(25 - 9/25) = 2√154/5
Area of △ ABC: | ab | * H / 2 = 2 √ 154 / 5 * 3 / 5 * (1 / 2) = 3 √ 154 / 25
The formula of encounter problem?
In the encounter problem, there are the following relations among total distance, encounter time and speed sum: ① speed sum × encounter time = total distance; ② total distance △ speed sum = encounter time; ③ total distance △ encounter time = speed sum
When x = (), the algebraic formula 2 (3x + 1) and 3-x are opposite to each other
2(3x+1)=-(3-x)
6x+2=x-3
5x=-5
x=-1
-1
2(3x+1)+3-x=0 6x+2+3-x=0 5x=-5 x=-1
What is the formula for solving the meeting problem?
Encounter time = distance and speed and
Speed sum = distance and △ encounter time
Distance sum = speed sum × encounter time
Known: polynomial X & # 179; + 3x & # 178; + ax + B divided by X & # 178; + X-1, remainder is 2x + 1, find a sum
x³+3x²+ax+b=X(x²+x-1)+2(x²+x-1)+2X+1
=x³+x²-x+2x²+2x-2+2X+1
=x³+3x²+3x-1
a=3 b=-1
What is the formula of mathematical pursuit problem and encounter problem!
The one above is very good and concise. Let me elaborate on it in detail: pursuit problem: pursuit time = distance before pursuit / speed difference. Encounter problem: encounter time = distance before encounter / speed difference. For complex travel problems, try to solve them by drawing line diagram