A few simple quadratic functions Fill in the blanks! Urgent

A few simple quadratic functions Fill in the blanks! Urgent

Bounty Points: 50 - 14 days and 23 hours until the end of the question.

The image of the quadratic function y=x +4x-1 is open upward, its axis of symmetry is a straight line x=-2, and the vertex coordinates are 2,-5).

It is known that the function y=3x +6x-2, when x=-1, y has a minimum value, which is 6

When x is 1, y increases with the increase of x; When x is 1, y decreases as x increases.

If the highest point of the image of a quadratic function is p(1,4), then when x = 1, the function y has the maximum.

The large value is 4; The axis of symmetry equation is x=1 and y increases as x increases when x is 1.

Big; When x is 1, y decreases as x increases.

It is known that the parabola y=-x + (2m-1) x-m +1 (m is constant) 1When m = 1, the parabola passes through the original.

Point 2The vertices of the parabola are on the y-axis when m= 3When m=, the vertex of the parabola.

On the x-axis 4When m, the axis of symmetry is to the left of the y-axis.

The coordinates of the intersection of the parabola y=x -x-12 and the x-axis are (0) and the distance between the two intersections is equal to 7

That's all、、、

The image of the quadratic function y=x +4x-1 is open upwards and its axis of symmetry is the straight line x=-2

The vertex coordinates are (-2, -5).

It is known that the function y=3x +6x-2, when x=-1, y has a minimum value, which is -5; When x is 1, y increases with the increase of x; When x is 1, y decreases as x increases.

If the highest point of the image of a quadratic function is p(1,4), then when x = 1, the function y has the maximum.

The large value is 4; The axis of symmetry equation is y=a(x-1)+4; When x is 1, y increases as x increases.

Big; When x is 1, y decreases as x increases.

It is known that the parabola y=-x + (2m-1) x-m +1 (m is constant) 1When m = plus or minus 1, the parabola passes through the original.

Point 2When m=1/2, the vertex of the parabola is on the y-axis 3When m = 5/4, the vertex of the parabola.

On the x-axis 4When m is 1/2, the axis of symmetry is to the left of the y-axis.

The coordinates of the intersection of the parabola y=x -x-12 and the x-axis are (4,0)(-3,0), and the distance between the two intersections is equal to 7

Easy to obtain: The coordinates of point c are (0,c).

Let the coordinates of the intersection point with the x-axis be x1, x2, (x1

So: c 2 = the child is grimmed and throbbed and the river contains Lou|x1||x2|=-x1x2=-c a, (can be obtained by triangle similarity.) )

So, c = -1 a

ac=-1

The triangle ABC is.

In a right-angled triangle, then the angle c is a right angle, the two points a and b are symmetrical with respect to the y-axis, and ac=ab, then the distances from the points a, b, and c to the origin are equal.

So the equation ax 2 + bx + c = 0.

If the numbers are inverse to each other, then b=0, AC heterogene.

The intersection point with the y-axis is, when x=0, y=c

The intersection point with the x-axis is when y=0, x=(c a).

square root, so: c square = c a

then: ac=-1

y=x^2-2(m-1)x+m2-2m-31.Proof:

[-2(m-1)] 2 - 4 1 (m2-2m-3)= 16 > 0 Constant True.

So, regardless of the value of m, the image of this quadratic function must have two intersections with the x-axis2Remembering a(x1,0), b(x2,0), Veda's theorem: x1+x2 = 2(m-1); x1x2 = m^2-2m-3

1/x1+1/x2 = (x1+x2)/x1x2 = 2(m-1)/(m^2-2m-3) = 2/3

m = 0 or 5

y=x 2+2x-3 or y=x 2-8x+12

(1).Discriminant 4(m-1) 2-4(m 2-2m-3)=8>0, so the function image and the x-axis have two intersections.

y= -2x^2+8x=-2(x^2-4x+16-16)=-2(x-4)^2+32

Maximum = 32

y=5x 2+6x-3=5(x 2+3 5x+9 25-9 beats only rotten 25)=5(x 2+3 5) 2-24 5

Minimum = -24 5

y=x+1-x 2=-(x 2-1 2x+1 4-1 4)+1=-(x-1 2) 2+5 Shanmo 4

Maximum Leak = 5 4

You rookie, you don't understand this.

>=8b

b^>=a

b=1, a>=3, false.

b=2,a>=4,2^2>=a;

So a + b = 16 + 4 = 20

Under the root number (b 2 + 16) = 10 = " b = 3 or -3) 3y=-x +2x+3 or y=x -2x-3

Because the vertex coordinates are (3, 2).

So the axis of symmetry is x = 3

Because the intersection of the image with the x-axis is symmetrical with respect to the axis of symmetry, and the distance between the two intersections is 4, the coordinates are (3 - 2, 0) (3 + 2, 0), i.e. the coordinates are (1, 0) and (5, 0).

The quadratic equation can be y=a(x-3) 2-2

From the symmetry, it is easy to obtain that the abscissa of the midpoint of the intersection point of the image and the x-axis is 3, then the left point is (1,0) and the right point is (5,0).

Bring in to get a = 1 2

As you can see, the intersection points with the x-axis are (1,0),(5,0), and the expression of this function is y=x2-6x+5. This is sought from the conditions you give. But I don't think the last question you asked seems right.

The hypotenuse is 2x, and the right-angled edge is the root number 2x, so y = half of the square of the root number 2x, i.e. y = x 2, x > 0

1.y=3, a(2,3)--symmetrical point change-y,(2,-3)2.c should be a range: the opening is facing up, and the absence of intersection on the x-axis means that y-always>> 0, y>(x-2) 2, c>4

3.??4b+b=0?b=0?

0),q(-1,0),pq=2

8+2x)*(60-3x)=-6x^2+120x-24x+480=-6x^2+96x+480

6 (x 2-16x-80) x = 8, grade 9.

y=x -x+1, the minimum value is 3 4, and there is no intersection with the x-axis.

y=-2x -(3 roots, number, 2)x, the maximum value is 9 4 has an intersection with the x-axis.

Draw, take the extreme, no intersection.