Solve the junior 3 quadratic function math problem

The two-digit solution of ls is wrong, the axis of symmetry is x=-b 2a, not -2b a or -b a, the meaning of this problem is to take the verification, point a coordinates, and point b coordinates as known conditions, and the function analytic formula as an unknown condition.

y=ax^2+(2-2a)x+a

The axis of symmetry is x = -b 2a = a-1 a = 2a = -1 analytically .

y=-a^2+4x-1

As for the second question, just find a random point in the analytic formula, such as (3,2).

1> over (0a) to get c=a(12) to get a+b+c=2The axis of symmetry yields -2b a=2 and gives a=c= from the three formulas

2> Take any abscissa point C.

Hehe, this question is a mistake!!

Because: over (1,2), substitution: a+b+c=2 and (0,a), substitution: c=a, linked to the above formula: 2a+b=2, that is: b=2-2a

Therefore, the original formula can be written as: y=ax2+(2-2a)x+a, so the axis of symmetry is x=-b a=-(2-2a) a=2-2 a, so that the verification is x=2, that is, 2 a should be equal to zero, which is not logical at all.

What do you think?

y=(mx-2m-2)(x-1)

So the abscissa is (2m+2) m and 1

2m+2) m=2+2 m is a positive integer.

then 2 m is an integer.

So m= 2, 1

Then 2+2 m=3,1,0,4

where x=1 and 0 are rounded.

So m = 2 or 1

So y=2x -8x+6 or y=x -5x+4

Proof: (1) If m=0, then the equation is -2x+2=0, and there must be a real root (2) If m≠0, then = -(3m+2) 2-4m*(2m+2)=m2+4m+4=(m+2)2

Regardless of the value of m, (m+2)2 is greater than or equal to 0, i.e. 0 The equation has a real root.

synthesis (1) (2), so that no matter that m is any real number, the equation has a real root.

Quadratic function y= m x2 - (3m + 2) x + 2m + 2 = (m x - (2m + 2)) (x-1).

Therefore, the abscissa of the two intersections with the x-axis is 1, (2m+2) m=2+2 m, because the abscissa is a positive integer, so m can only be -2 (rounded, at this time, x is 1, which does not meet the two intersection points), 1, 2;

So the quadratic function is y=x2-5x+4 or y=2x2-8x+6

mx -(3m+2)x+2m+2=0, when m≠0 the value of the discriminant (3m+2) -4m(2m+2)=(m+2) so x1=(m+2) m, x2=(3m+2-m-2) 2m=1. So when the integer m=1, 2. When m=1, y=x -5x+4.

When m=2 is y=2x -8x+6.

It is known that p1(x1,2001) p2(x1,2001) is two points on the image of the quadratic function y=ax2+bx+7, try to find the value of the quadratic function when x=x1+x2.

Analysis: Quadratic function y=ax2+bx+7

When x=0, y=7

p1 (x1, 2001) p2 (x1, 2001) two dots on its image.

x1, x2 are symmetrical with respect to the point x=-b 2a.

When x1<00, -b 2a>0==>b<0x2+x1 is symmetrical with 0 about the point x=-b 2a.

The quadratic function has a value of 7 when x=x1+x2

When 0x1, x2 is symmetrical about the point x=-b 2a.

x2+x1=-b/a

x=-b a and x=0 are symmetrical about the point x=-b 2a.

The quadratic function has a value of 7 when x=x1+x2

The values of x1, x2, and y are equal, and when x=0, y=7, then the parabolic opening is facing up, x1+x2=t, t is symmetrical with x=0, then when x=x1+x2, y=7

p1(x1,2001) p2(x1,2001) is a quadratic function y=ax2+bx+7 two points on an image.

> x1 and x2 are symmetrical with respect to the straight line x=-b (2a).

>x1+x2=-b/a

>f(x1+x2)=f(-b/a)=b^2/a-b^2/a+7=7

> when x=x1+x2, f(x)=7

From the inscription, it can be seen that a is 1

So the equation becomes y=x2+bx+cThe key to solving the problem is the area formula.

The area is equal to 1 2*ab*4c-b

Really? But AB is another problem, so to transform, using Veda's theorem, AB can be written under the root number.

x1+x2）2-4x1x2

It's under the root number.

B2-4C can be brought underneath.

The area is equal to 1 2 * (b2-4c under the root number) * 4/4 c-b = 1, and you should be able to find it.

The steps on the first floor are correct, just a little careless.

2(k+1)≠0

k≠-1y=2(k+1)x²+4kx-2k-3

b²-4ac=16k²-4(2(k+1))(2k-3)>02k²+(k+1)(2k+3)>0

2k²+2k²+5k+3>0

4k²+5k+3>0

b²-4ac=25-48<0

k is any real number other than -1.

Solution: (1) The parabola has two intersection points with the x-axis, 0, (4k)2-4 2(k+1)(2k-3) 0, arranged, k+3 0, solved, k -3

Therefore, when k -3, the parabola has two intersections with the x-axis.

Knowing the parabola y=2(k+1)x2+4kx+2k-3, find:

1) What is the value of k, when there are two intersection points between the parabola and the x-axis?

2) Why is there no intersection between the parabola and the x-axis when the value of k? Test Point: The Intersection of the Parabola and the X-Axis Topic:

Calculation Problem Analysis: According to the relationship between the quadratic function and the unary quadratic equation, the intersection problem of the parabola and the x-axis is transformed into the discriminant formula of the root, and the inequality solution is listed Answer: Solution:

1) The parabola has two intersections with the x-axis, 0, (4k)2-4 2(k+1)(2k-3) 0, finished, k+3 0, solved, k -3

Therefore, when kk -3, the parabola has two intersections with the x-axis.

2)) There is no intersection between the parabola and the x-axis, 0, (4k)2-4 2(k+1)(2k-3) 0, finished, k+3 0, solved, k -3

Therefore, when k -3, the parabola and the x-axis have two intersection points Comments: This question not only examines the relationship between the quadratic function and the unary quadratic equation, but also examines the discriminant formula of the root of the unary quadratic equation

1.Solution: From y=x -x+m, we can know that y=(x-1 2) +m-1 4).

Therefore, (1) the opening direction is upward, the axis of symmetry is x=1 2, and the vertex coordinates are (1 2, m-1 4).

2) When m>1 4, the vertex is above the x-axis;

When m=1 4, the vertex is above the x-axis;

When m<1 4, the vertex is below the x-axis.

2.Solution: According to the problem, the parabolic equation y=-1 4x +4 constructs the coordinate system, and the rectangle is also put into the coordinate system, then the coordinates of the four vertices in the square are (-4,-2),(4,-2),(4,0),(4,0) respectively

1) When the tunnel is a one-way street, the truck can drive in the middle of the tunnel, and the positions of the vertices on both sides of the truck in the coordinate system are (-1,2) and (1,2).

At x = -1 and 1, the coordinates on the parabola are (-1,4-1 4) and (1,4-1 4), respectively

The height corresponding to the change point is: 2+(4-1 4)=

That is, they are all higher than 4m, so they can pass through the tunnel.

2) If there is a double lane in the tunnel, change the truck to study the central axis, the central axis out of the tunnel is 6m high, the x-axis coordinate value at the other side is 2, and the corresponding parabolic ordinate value y=4-(1 4)*2 =3, that is, the tunnel height here (2+3)m=5m>4m

Explain that the truck can still pass through the tunnel at this time.

1.(1) Opening direction: up Axis of symmetry: x= vertex coordinates (, m-1 4).

2)m>1\4 m=1\4 m<1\4

2.(1) and (2) can pass this question, I just did it in the morning

(2008 Guigang) It is known that the roots of the two real numbers of the unary quadratic equation x2-4x-5=0 are x1 and x2, and x1 can send the graph over and the coordinates of the key points :(1) are obtained from the equation x2-4x-5=0.

According to the meaning of the title, it is a quadratic function, because the corresponding g and r are fixed values, there is only one independent variable, and the exponents are quadratic, so it is a quadratic function of one variate, so b is chosen.