Problems of quadratic functions and binary linear equations

1) There is only one intersection point between y=x 2-2x+2m and y=mx, which means that the equation x 2-2x+2m=mx has a double root, then the discriminant formula =(-m-2) 2-4*2m=0 gives m=2.

2) The vertex coordinates of the parabola y=x 2-2ax+1 are (a, -a 2+1), so -a 2+1=2a+1 , and the solution is a=0 or -2.

3) The two equations are conjoined to obtain x 2-2x+5-b=0, a) the image has two intersections, that is, the equation has two different real roots, so the discriminant formula = 4-4(5-b)>0 , and the solution gives b>4.

b) The image has only one intersection point, that is, the equation has two equal real roots, so the discriminant formula = 4-4(5-b)=0 and the solution is b=4.

c) If the image has no intersection point, the equation has no real root, so the discriminant equation = 4-4(5-b)<0 , and the solution is b<4.

d) The image has at least one intersection point, that is, the equation has at least one real root, so the discriminant equation = 4-4(5-b)>=0 , and the solution is b>=4 .

m=2a=0 or -2

The image has two intersections, that is, the equation has two different real roots, so the discriminant formula = 4-4(5-b)>0 , and the solution is b>4.

The image has only one intersection point, that is, the equation has two equal real roots, so the discriminant formula = 4-4(5-b)=0 , and the solution is b=4 .

If the image has no intersection point, the equation has no real root, so the discriminant formula = 4-4(5-b)<0 and the solution b<4 is obtained.

The image has at least one intersection point, that is, the equation has at least one real root, so the discriminant equation = 4-4(5-b)>=0 , and the solution is b>=4 . Appraise.

m=6+4 2 or 6-4 2 a= two intersections: b>4 a: b=4 none: b<4 at least one: b 4

m=2a=0

There is no intersection at 4>x.

4=x with only one intersection.

There are two intersections at 4>x.

Quadratic functions withUnary quadratic equationsA specific definition means that a quadratic function has a basic expression, and the quadratic function must be quadratic if it is the highest degree of x, and the image of the quadratic function on the coordinate axis is a symmetry axis parallel to the y-axis or coincides with the y-axisParabola

So a quadratic function is a definition of this. However, the relevant definition of a quadratic equation is that it is an integer on both sides of the equal sign.

Moreover, Liangcha only contains one unknown, that is, it only contains an unknown like x, and then the highest number of times that Lu Hui knows is two, and the highest number of times that Lu Hui then knows this is the definition of two, which is basically the same as the quadratic function.

So the above is a quadratic function and a quadratic function.

Definitions.

A quadratic equation is an equation in which there is only one unknown, and the number of unknowns is quadratic, and the solution of this equation has two modes, and there are different ways to solve the equation.

A quadratic function is a functional relation that changes with the change of x, and a quadratic equation is just a special point of the quadratic function, that is, when the value of the quadratic function is 0, its relation is a quadratic equation.

Quadratic function and unary quadratic equation relationship:

1. From the point of view of form:

Quadratic function: y=ax bx c (a≠0).

Unary quadratic equation: ax bx c=0 (a≠0).

2. From the content point of view:

A quadratic function represents the relationship between a pair (x,y) with an infinite number of state pairs; A quadratic equation represents the value of the unknown number x, with a maximum of 2 values.

3. Mutual relationship:

The abscissa of the intersection of the quadratic function and the x-axis is the root of the corresponding unary quadratic equation.

For example, if the intersection of y=x 4x 3 and the x-axis is (1,0) or (3,0), then the root of the unary quadratic equation x 4x 3=0 is x=1 or x=3

Algebraically speaking:

A quadratic equation is an unknown problem with two answers, sometimes degenerating into a virtual state with one answer, or no answer (no solution).

A quadratic function is two unknowns, or two variables, and a quadratic function refers to their correspondence. One of the variables gives a value, and the other variable can have two corresponding values, or one, or none. The former is called the dependent variable and is usually denoted by y; The latter is called an independent variable and is usually denoted by x.

The range of values for the independent variable is called the definition domain, and the range of values for the dependent variable is called the range of values (of the function). In general, there is no restriction on defining the domain, and there is a limit to the range of values for any quadratic function.

For a given function value, the quadratic function degenerates into a unary quadratic equation.

Analytic geometry:

The quadratic function is a parabola, and its general form family is y = ax 2 + bx + c, when a>0 it is a parabola with an opening upward, and when a < 0 it is a parabola with an opening downward. A quadratic function may have two intersections with the x-axis, may have a tangent point, may have no intersection point, and may not have a shear mask. When y = 0, the intersection of the quadratic curve y = ax 2 + bx + c with the x-axis becomes the solution of the quadratic equation ax 2 + bx + c = 0.

If a quadratic equation ax 2 + bx + c = 0 has a solution, its solution is fixed, i.e. the intersection of the quadratic curve y = ax 2 + bx + c depicted by the quadratic function with the x-axis is fixed. However, the quadratic function can be changed, i.e., there can be an infinite number of quadratic curves passing through two fixed points on the x-axis.

x (0, x1) means that x belongs to the (0, x1) range, i.e., 0

That's the closed interval, which is the open interval, which means 0 is less than x and less than 1

This means that x is within the range of (0,x1).

This is an open interval *content* that can be understood as well as a set.

Let f(x)=ax 2-2ax-1=a(x-1) 2-(a+1) ax axis of symmetry be x=1, so f(x) is monotonic f(2)=-1<0 on (1, 2).

f(1)=-(a+1)<=0 => a>=-1, so the value range of a is a>=-1

ax -2ax-1=a(x-1) -a+1) ax of symmetry is x=1

f(1)=-(a+1) is the maximum.

a+1)≤0

Therefore, the value range of a is a -1

1.=b -4ac=(2m-1) -4*1*(m-m-2)

Therefore, the parabola and the x-axis must have two different points of intersection.

2.Let y=0, then we get:

x²-(2m-1)x+m²-m-2=0

Use the formula method to solve the 1 element quadratic equation (m, where it is considered to be a constant. ）xa=(-b+√△/2a=(2m-1+3)/2=m+1xb=(-b+√△/2a=(2m-1-3)/2=m-23.According to the title, let the coordinates of point c be (x,y) then:

y=x²-(2m-1)x+m²-m-2 1|xa-xb|=|m+1-m+2|=3 The area of 2 abc is equal to |xa-xb|*|y|=3*|x²-(2m-1)x+m²-m-2|=6 3

Do the rest yourself... My brain is dizzy, hahaha

Only 5 points for two questions? There will be at least 40 points for these two questions during the exam, and it is not difficult to do, but it is not very good for you to send this whole question up for others to do with you.

1△=9>0

So the parabola and the x-axis must have two different points of intersection.

2.Let y=0, then we get:

x²-(2m-1)x+m²-m-2=0

Solve this equation to obtain: xa=(-b+ 2a=(2m-1+3) 2=m+1

xb=(-b+√△/2a=(2m-1-3)/2=m-23.According to the title, let the coordinates of point c be (x,y) then:

y=x²-(2m-1)x+m²-m-2 1|xa-xb|=|m+1-m+2|=3 The area of 2 abc is equal to |xa-xb|*|y|=3*|x²-(2m-1)x+m²-m-2|=6 3

Quadratic function y ax 2+bx+c

The unary quadratic equation ax 2 + bx + c 0

So it is obvious that when the function value of the quadratic function is y 0, the quadratic function becomes a quadratic equation, and the solution of the quadratic equation is the abscissa of the intersection point of the quadratic function and the x-axis, that is, the zero point of the quadratic function.

That's the first question.

The opening is upward, delta < 0

Is y=(2m-1)x 2+(m+1)x+(m-4)?

delta=m^2+2m+1-8m^2-36m+16<0=>-7m^2-34m+17<0

When it is equal to 0.

m>17 7 o'clock.

Too sleepy, not necessarily right.

Because the function curve is too good at (m,0) and (n,0), it is judged that m 2-2007*m+2008=0 and n 2-2007*n+2008=0;Therefore, the result of algebraic matching is 0+0+2009=2009