The quadratic function y x 6x 8 is known

1) Let y=x-6x+8=0, i.e. (x-2)(x-4)=0, x=2, or x=4, and the intersection points with the x-axis are (2,0) and (4,0).

Let x=0, then y=x -6x+8=8, and the intersection point with the y-axis is (0,8)2)y=x -6x+8=(x-3) -1, vertex coordinates.

for (3,-1).

3) Let y=x-6x+8=(x-2)(x-4)<0, 2 axes of symmetry.

is x=3, when x 3, y decreases as x increases. Hope.

For the image of the quadratic function y=ax 2+bx+c:

1. Open a>0 up, a<0, down.

2. The axis of symmetry x=-b 2a

3. Vertices [-b 2a, (4ac-b 2) 4a], the ordinate value of this point is the extreme value of the function.

4. The intersection point with the x-axis, the value of the discriminant formula B 2-4ac (no intersection point when it is less than 0, and when it is greater than 0, two intersection points, equal to one intersection point when 0) 5. The intersection point with the y-axis (0, c).

6. When the opening is upward (a>0), the function value y on the left side of the axis of symmetry decreases with the increase of x, and the function value on the right side increases with the increase of x. The situation of a<0 is the opposite of the previous one.

1) The intersection point with the x-axis is at y=0, that is, x-6x+8=0, and the equation is solved to obtain x=2 or x=4;So the coordinates of the intersection point of the image with the x-axis are (2,0) and (4,0). The intersection point with the y-axis is at x=0, i.e., y=0-0+8=8;So the coordinates of the intersection of the image and the y-axis are (0,8).

2) From the first problem, the minimum value of y is y=-1 when x=3 (this is by using symmetry, and another method is by using the evaluation formula), then the vertex coordinates of the image are (3, -1).

3) Because a is greater than zero, the function image is upward, and because when x = 2 or x = 4 the function value y=0, when x is greater than 2 and less than 4, the function value is less than zero (using the threading needle method).

4) Because a is greater than zero, the function image is upward, and the minimum value of y is y=-1 when x=3, so when x is less than 3, y decreases with the increase of x (taking advantage of the monotonicity of the function, the number can be combined to help understanding).

1, =8*8-4*3*4>0, so there are two intersections with the x-axis.

2. The root of the equation is the horizontal sitting of the intersection of the function and the x-pumped point.

3, when y=-1, x=1 or 5 3

Hope I can help Sakura talk about you!

y=ax +bx+c=x -6x+9, and its discriminant formula:

b²-4ac=（-6）²-4×1×9=36-36=0。

Therefore, Changsong Cave, the equation x -6x + 9 = 0 is a pair of real numbers such as the root of the real number of the withering. The image of its function is:

y=x²-x-6=0

x-3)(x+2)=0

x=3 or x=-2

So the intersection point with the x-axis is (3,0) (2,0).

Let x=0 then y=-6

So the intersection point with the y-axis is (0,-6).

The ** of 2 x 32 questions is not easy to draw 、、、

1.The coordinates of the intersection of the quadratic function image and the coordinate axis: with y-axis a(0,-6) and x-axis b(-2,0) c(3,0) vertices p(1 2,-23 4)2Look at the picture.

3.It is pointed out that the solution of the equation x -x-6=0 x1=-2 x2=3 inequality x -x-6<0 is valid in the range of x -24 s△abc=1/2*|-6|*|cb|=15

This chapter is the focus.

yx²+4x+4-4

Equipped with a completely flat method. Complement the constants.

x+2)²-4

So. The axis of symmetry is.

x=-2 vertices are (-2, -4).

The intersection of the function and x.

In fact, let y=0 find 0

x +4x solution. So x1

The two intersections of x2 strands are: (0,0).

The image of the quadratic function y=x2+ax+a-2 has two intersections with the x-axis, which is equivalent to the equation.

x 2+ax+a-2=0 has two different real roots.

Let the intersection points be (x1,0),(x2,0).

Then x1, x2 are two different real roots of the equation x 2 + ax + a-2 = 0.

Then the distance between them =|x1-x2|= root number down-slap [(x1+x2) 2-4x1x2].

According to Veda's theorem.

x1+x2=-a

x1x2=a-2

So under the root number [(x1+x2) 2-4x1x2] = under the root number (a 2-4a+8) = 2 times the root number of the Huai chain 5

a^2-4a+8=20

a^2-4a-12=0

a=-2 or a=6

=（x-1/2）^2+m-1/4

Therefore, the image opening of the quadratic function is upward.

The axis of symmetry is x=1 2

The vertex coordinates are (1 2, m-1 4).

2. The image has an intersection with the x-axis.

The vertices of the quadratic function are below or on the x-axis.

Therefore, the longitudinal mark of the vertex m-1 4 0

i.e. m 1 4

1.If x and y are set to 0, the coordinates of the intersection points are (0,8), (2,0), (4,0), respectively, and (3,-1) at the vertex

2.Draw the picture yourself.

a, the solution is 2 and 4

The set of solutions where b is greater than zero is handed over.

The set of solutions where c is less than zero is.

1) Find the vertex coordinates of the quadratic function image and the intersection coordinates (2) Draw the function image (3) Observe the image and point out the solution of the equation x -x-6 = 0 and the range of x values that make the inequality x -x-6<0 hold (4) Find the area of the triangle formed by the intersection of the quadratic function image and the coordinate axis.

1) Let y=0

x²-x-6=0

x1=-2, x2=3

Quadratic function image and x-axis intersection coordinates (-2,0), 3,0)y=x -x-6=(x-1 2) 2-25 4 vertex coordinates (1 2,-25 4).

2) Draw a function image --- sketch.

3) Observe the image and point out the solution of the equation x -x-6 = 0, that is, the coordinates of the intersection point between the quadratic function image and the x-axis.

x1=-2, x2=3

The quadratic function image, with the opening pointing upwards, makes the part of the inequality x -x-6<0 below the x-axis take the value range of x: -2(4) Find the area of the triangle formed by the intersection of the quadratic function image and the coordinate axis so that x=0

y=x²-x-6=-6

The intersection of the quadratic function image with the y-axis (0,-6).

Area of the triangle = ab*oc 2 = 5 * 6 2 = 15