A math problem for the third function of junior high school if it is good, it will definitely give

Because 1 (x1) +1 (x2) = (x1 +x2 ) (x1x2) =(x1+x2) -2x1x2 (x1x2) =2

That is, (x1+x2) -2x1x2=2(x1x2) and the function y=(m-1)x +(m-2)x - the two intersections of 1 and the x-axis are a(x1,0) and b(x2,0).

So x1+x2=(2-m) (m-1), x1x2=-1 (m-1), so (2-m) (m-1) +2 (m-1)=2 (m-1) 2-m) +2(m-1)=2

So m=0 or 2

Because =(m-2) +4(m-1)>0

So m >0

So m=2

Machining the second equation yields (x1+x2) 2-2x1x2 (x1x2) 2=2

When y=0 there are two intersections, x1x2=-1 m-1 x1+x2=-(m-2) (m-1).

Bring it in and do the math yourself, pay attention to the value

Substitute y=0 and x=0 respectively to solve a, b, and c, and use the abscissa of a and b to find the axis of symmetry.

2) Solution: Let the relation of the primary function bc be y=kx+b, and substitute the coordinates of b and c.

0=3k+b,3=b, and k=-1,b=3 is a primary function relation

Replace p(m,? Substitute y=-x+3 to get y=-m+3. i.e. p(m,-m+3) pf de, f(m,? ）

Replace f(m,? ) is substituted by y=-x +2x+3 to get y=-m +2m+3. i.e. f(m,-m +2m+3).

pf=-m +2m+3-(-m+3)=-m +3m If the quadrilateral pedf is a parallelogram, because pf de, so pf=de substitutes x=1 into y=-x+3 to get y=2, so e(1,2) and d(1,4), de=2, that is, -m +3m=2 solves m1=1 (on the axis of symmetry, does not meet the topic, rounds off), m2=2 pf=-m +3m, when m=2, the quadrilateral pedf is solved by the parallelogram itself The process can be a bit cumbersome.

Hope it helps.

b2-4ac<0, the function has no intersection with the x-axis, right?

(1) Multiply the crosses and simplify them to get x1=(x+3)(-x+m+1) to get x1=-3 x2=m+1 Bring x=0 into the ordinate of point C: 3m+3 according to OA +ob =2oc+1 bring in (-3) +x+m+1) =2(3m+3) to get m1=3 m2=1 because x1 x2 so m=1

The analytic formula is y=—x -1x+6

2) c(0,6) let y=kx+b bring the point c into to get y=kx+6, the two equations have an intersection point, and the simultaneous —x -1x+6=kx+6 is sorted out to obtain -x -(1+k)x=0

There is only one intersection point, so (dell it) = 0 i.e. (1+k) =0 so k = -1

So the straight line is y=-x+6

The answer is this: (1) m-3≠0 m≠3 x1+x2=-2m m-3 and x1+x2≠0

2m／m-3≠0∴m≠0

There are two unequal real roots.

＞0∴b²-4ac＞0

4m²-4（m-3）（m+1）＞0

m＞-2／3

m -2 3 and m ≠ 0 and 3

2) When m=2, m is the smallest even number.

Substitute m=2 into the evaluation.

Classmates, substitute yourself later.

It's simple.

It's simple.

As long as the simultaneous inequality m≠3

2m≠0 is enough, and the rest of the problems are not difficult.

(1) The equation is a quadratic equation: m-3 is not equal to 0

The equation has two unequal real roots: it shows that the discriminant formula is greater than 0 and these two roots are not opposite to each other: it shows that 2m m-3 is not equal to 0 (Vedic theorem) (2) The range of values of m can be found from (1), and the smallest even number can be taken and brought into the original equation to find the roots.

(1) From the meaning of the question: =4m 2-4(m-3)(m+1) 0, solve m by the root of the equation x1+x2=-2m (m-3) from the Vedic theorem, since the two roots are not opposite to each other, so -2m (m-3) ≠ 0, that is, m≠0 and m≠3

Therefore, the range of m is m and m≠, m≠3

2) Since the range of m is m and m≠, m≠3, so m=0 at this time, so the equation becomes -3x 2+1=0

x1 = 3 3, x2 = - 3 3

The coefficient of the quadratic term of the quadratic function is greater than 0, so the left side of the axis of symmetry gradually decreases and the right side gradually increases in the r range, and the minimum value is obtained at the axis of symmetry.

axis of symmetry x=-a2

1) At that time, -a 2<<-1

Therefore, the minimum value at -1 is taken as ymin=1-a+3=4-a

1) A(1,0)B(0,2) After the rotation of a1(0,-1)b1(2,0), let y=kx+b point be brought in.

2) Let y=ax2 (x squared by a times) + bx + c

Three-point coordinates are brought in OK

1) y= (2) The square of y= is too simple to talk about.

Solution: (1) a b is the intersection of y=-2x+2 with the x-axis and y-axis, a(1,0) b(0,2).

a1ob1 is rotated 90° counterclockwise by oba around the point o to obtain a1(0,1) b1(-2,0) and brought into y1=kx+b to obtain y1=.

2) Let the functional relation of the parabola with a, a1, and b1 be y2=ax 2+cx+d

Bring in the three points a, a1, and b1 to get y2=

This 1a1 (0,1) b1(-2,0).

The new straight line expression is y=(1 2)x+1

2. Let the equation of the parabola be y=ax squared + bx+c, bring a, a1, and b1 into the solution, and solve a=1b=1c=1

y=x+x+1

It's very simple, find two points and then find the symmetry point, then substitute the function formula once, find the two points, and then set the parabola, and bring in and find out.

y=ax^2+bx+c

When y=0, x has two values, so y=a(x+2)(x-x1)=ax 2+a(2-x1)x-2*a*x1

So b=a(2-x1), c=-2*a*x1 and the intersection of the positive semi-axis of the y-axis are below the point (0,2). It means that the opening is downward, a<0;

Because 10, 1 pair.

2a+c=2a(1-x1)>0,2 pairs;

4a+c=2a(2-x1)<0,3 pairs;

According to y=a(x+2)(x-x1), the intersection point with the positive half axis of the y-axis is below the point (0,2).

When x=0, y=-2a*x1<2, so 0>a*x1>-12a-b+1=1+a*x1>0,4 pairs;

The image of the primary function y=kx + five-twoths and the inverse proportional function y=x/m intersect at two points a(a,2)b(-4,m).

1.Find the analytic expressions of the primary function and the inverse proportional function.

y=m/xm=2a=-4m

a=-2my=kx+5/2

2=ak+5/2

m=-4k+5/2

2-m=(a+4)k

k=(2-m)/(4-2m)=1/2

m=-2+5/2=1/2

a=-1m=-2

y=m/x=-2/x

y=kx+5/2=x/2+5/2

2.Find the area of aob.

a(-1,2)

b(-4,1/2)

y=x/2+5/2

x=0y=5/2

y=0x=-5

The intersection point of y=x 2+5 2 and the coordinate axis is d(0,5 2)c(-5,0).

Area of AOB = Area of COD - Area of AOD - Area of COB = (5*5 2) 2-(1*5 2) 2-(5*1 2) 2=25 4-5 4-5 4=15 4

From the meaning of the title, we can get y1=k1 (1 x), i.e., y=k1x, y2=k2x, because y1+y2=y

So k1x+k2x 2=y

Substitute x=-1 y=0.

k1+k2=0

So choose C

From the meaning of the title, y1=k1x, y2=k2x 2, so y=k1x+k2x 2, substituting x=-1, y=0 can get: 0=-k1+k2, so choose c.

It is necessary to fully understand the method of substitution.

From the title: y1 = k1x y2 = k2x square.

Substituting -1 and 0 gives -k1+k2=0 but there is no such option.