Junior high school math problem 10

Eh: I can't upload the picture here.

Probably write an idea and do the math yourself.

3.The axis of symmetry x=-a2 a0, so x0 has two results: -1 -a, 2<0, then f(-2 a)=0, f(1)=-4

a 2<-1 then f(-1)=0 f(1)=-44 axis of symmetry x=-a 5The axis of symmetry is x=t

According to the above method, a,t are classified as the following categories where the definition fields of a and t are r, so consider all when considering the scope.

Secondly, when classifying, you can draw a picture, and then ask me if you don't understand.

I couldn't figure it out in the middle of the night2

1 m=4 vertices on the y-axis [(m-4) 2=0]m=2 or 14 vertices on the x-axis (m-4) squared =4(2m-3)m=3 2 vertices at the origin (2m-3=0).

2 l squared 16, i.e. (l 4) squared.

He's right, just count it that way.

Answer every one, that's too time-consuming. Why don't you rely on me to tell you.

1 m=4 vertices on the y-axis [(m-4) 2=0]m=2 or 14 vertices on the x-axis (m-4) 2=4(2m-3)m=3 2 vertices at the origin (2m-3=0).

2 l^2/16

1. Solution: If the value range of the independent variable is all real numbers, then there is a maximum value of the quadratic function. To put it simply, there is a minimum value for the opening upward; The opening has a maximum value downward.

Observe this quadratic function first, the value range of the independent variable is not particularly explained, then the default is the whole real number, and the coefficient of the quadratic term is greater than zero, and the opening is upward, then there is a minimum value, and this minimum value is actually the ordinate of the vertex. From the general formula y=ax +bx+c of the quadratic function, we can see that its vertex coordinates are (-b 2a, 4ac-b 4a), and the minimum value of this quadratic function is 5 by replacing a, b, and c with specific numbers

In this case, when the range of independent variables is given, the opening direction and the axis of symmetry are considered. If the opening is upward, then to the left of the axis of symmetry, the value of the function decreases as the independent variable increases; On the right side of the axis of symmetry, the value of the function increases as the independent variables increase. If the opening is downward, the left side of the axis of symmetry increases with the increase in the value of the function of the independent variable; On the right side of the axis of symmetry, the value of the function decreases as the independent variable increases. If the axis of symmetry is exactly within a given range, then the minimum value does not change if the opening is upward, and the maximum value is judged according to the range.

Specific to this problem, the axis of symmetry is x=-2, obviously in a given range, then bring in the minimum value of 5 in the function, and -3 is close to the axis of symmetry, and 3 is far from the axis of symmetry, then the maximum value should be obtained when x=3, and bring in y=55

2. In the analysis of the same problem 1, the image opening of the function is upward, and the vertex is (2,-1) obtained by the vertex coordinate formula, and the axis of symmetry is x=2,, between the given range [0,6], so the minimum value y=-1,0 is close to the axis of symmetry and 6 is far from the axis of symmetry, then the maximum value should be x=6 when y=15 is obtained

When x=-2, the function has a minimum value of 5

2.When x=-2, the function has a minimum value of 5, and when x=3, the function has a maximum value of 55

0≤x≤6)

When x=2, the function has a minimum value of -1; When x=6, the function has a maximum value of 15

1. y=2x^2+8x+13=2(x^2+4x+4)-8+13=2(x+2)^2+5

When x = -2 there is a minimum value of y = 5 and no maximum value.

If the value of the independent variable is -3 x 3.

There is a minimum value of y=5 for x=-2 and a maximum value of y=55 for x=3

There is a minimum value of y=-1 when x=2

There is a maximum value of y=15 when x=6

1. Because there is no limit to the value range of this function, according to a=2>0, the function opens upward, so there is a minimum value. According to c = 4ac-b 4a, so, ymin = 5, at this time, the function has a value range limit, because the symmetry axis of the function is -b 2a=-4 is not in the range of values, but it is on the right side of -3, so when x=-3, ymin=7;When x=3, ymax=55

2. This problem is similar to the previous question, first find the symmetry axis = 2 is within the value range, so, ymin=-4;When x=6, ymax=15

(1) Find the coordinates of the point A and the degree of AOB;

2) If the parabola y=1 2x2+2x is translated 4 units to the right, and then 2 units downwards are shifted to obtain the parabola m, and its vertex is the point c connecting OC and AC, and the AOC is folded along OA to get the quadrilateral ACOC Try to judge its shape and explain the reason;

3) In the case of (2), determine whether point c is on the parabola y=1 2x2+2x, please explain the reason;

4) If the point p is a moving point on the x-axis, try whether there is a point q on the parabola m, so that the quadrilateral with the vertices of o, p, c, q is a parallelogram, and oc is an edge of the quadrilateral? If it exists, please write the coordinates of the point q directly; If not, please explain why

Hard typing, hopeful, o(o haha.

1.By the nature of the function, k can only be less than 0, and then another =4-4k 0, we get k -12First, k +4k-5 0 i.e. k-1 k+5 0 gets k 1 or k -5

Another =16 1-k -12 k +4k-5 =4k -80k+76 0 gives 1 k 19

To sum up, yes. k -5 or.

1＜k＜19

1. Let y=kx -2x+k, because kx -2x+k is always negative, so the image of y is on the negative half axis of one axis, that is, the image opening of y=kx -2x+k is downward, and there is no intersection with the y-axis, so k<0, and (-2) =4-4k 0 -4k <0 so k<-1

2. Because the images of the function y=(k +4k-5)x +4(1-k)x+3 are above the x-axis, k +4k-5>0 and.

4(1-k)) 4*(k +4k-5)*3<0 so k -5 or.

1＜k＜19

The question is to test the properties of the function image, the quadratic term coefficient determines the opening direction, less than 0 down, greater than 0 up =b -4ac to determine whether there is an intersection point on the x-axis.

1. Because the original function is always negative, the function should be open downward and have no intersection with the x-axis.

In this case, two equations can be obtained: k<0, 2) -4xk <0 to get k>-1

2. The image is above the x-axis, indicating that the function opening is upwards and there is no intersection on the x-axis.

So there is, k +4k-5>0 , we get k 1 or k -5 = 16 1-k -12 k +4k-5 =4k -80k+76 0 , we get 1 k 19

In summary, we get k -5 or 1 k 19

1) If k=0, y=-2x cannot always be negative k≠02) If k≠0, then the given function is a quadratic function.

To make y=kx -2x+k constant negative for any real number x, the image opening is downward and has no intersection with the x-axis.

k<0 and 4-4k <0

Solution: k<-1

1) When k2+4k-5=0, k=-5 or k=1

If k=-5, then the images of y=24x+3 cannot all be above the x-axis, so k≠-5

If k=1, then the image of y=3 is above the x-axis.

2) If k2+4k-5≠0, then the given function is a quadratic function, and there should be k2+4k-5 0 and δ 0, i.e., (k+5)(k-1) 0, (k-1)(k-19) 0, and the solution is 1 k 19

From (1) and (2) 1 k 19

x=-(2) There is a maximum value of 1 k-2 k+k=k-1 k<0 at 2k, and -10Get k>1 or k<-5

3-16(1-k)^2/4(k^2+4k-5)>0.Get 1, so 1