The quadratic function y x 2 4x 3 is known

Solution: (1) y=x2-4x+3

x2-4x+4-4+3

x-2）2-1；

2) According to the quadratic function in (1).

The vertex relation of the function is (2,-1);

When x=0, y=3;

When y=0, i.e., x2-4x+3=0, the solution gives x=1 or x=3, the function image.

After the points (0,3), (1,0), (3,0);

3) From the image in (2), we can see that when 1 x 3, y 0

Because y=-x 2+4x-3

x^2-4x)-3

x^2-4x+4)+1

x-2)^2+1

So the function has a maximum value: 1, which corresponds.

Argument. The value is 2,2).

Solve equations. x^2+4x-3=0，x-1)(x-3)=0x1=1,x2=3

So. The coordinates of the intersection of the image and the x-axis of the function (1,0),(3,0).

The idea should be: point D is the farthest away from the straight line BC! Because in a triangular BCD, with BC as the base, the farther the distance, the higher the height, and the larger the area.

The maximum value is obtained at a fixed point:

Maximum: (4ac-b 2) (4a)=(4*3-16) (-4)=1 The value of the corresponding independent variable: x=-b (2a)=4 2=2x 2+4x-3=0

x^2-4x+3=0

x-3)(x-1)=0

x=3 or 1 intersection coordinates: (1,0); （3，0）

2) Steps to draw an image: list, dot, line;

3) When y 0, that is, the part of the image above the x-axis, then write the value range of x 1) y=x2+4x+3,y=x2+4x+4-4+3,y=x2+4x+4-1,y=(x+2)2-1;

2) List: x.... 4 -3 -2 -1 0 …

y … 3 0 -1 0 3 …The image is shown in Figure 3) From the image, it can be seen that when x -3 or x -1, y 0 Comments: Test points of this question: Three forms of quadratic functions; Image of a quadratic function Test Center Comments:

This question examines the analytic form of the quadratic function and the drawing of the parabola

1) General formula: y=ax2+bx+c(a≠0, a, b, c are constants);

2) Vertex formula: y=a(x-h)2+k;

3) Intersection (with x-axis): y=a(x-x1)(x-x2).

2) Steps to draw an image: trivial list, draw points, and connect lines;

3) When y 0, that is, the part of the image above the x-axis, write the value range of x 1) y=x2+4x+3, y=x2+4x+4-4+3, y=x2+4x+4-1, y=(x+2)2-1;

2) List: x.... 4 -3 -2 -1 0 …

y … 3 0 -1 0 3 …The image is shown in Figure 3) It can be seen from the image staring that when x -3 or x -1, y 0 1, the quadratic function y=x 2+4x+3 is known

1) Y=X 2+4X+3 is converted into Y=A(X-H) 2+K by matching method;

2) draw an image of this quadratic function in a plane right-angled buried coordinate system;

3) Write down when x is worth y0

When y=3, x2

4x+6=3, i.e. x2

4x+3=0, scrambled.

Solution: x11, x2

So the answer is: and Li Xun 1 or 3

1) Because y=-x 2+4x-3

(x^2-4x)-3

(x^2-4x+4)+1

(x-2)^2+1

So the function has a maximum value: 1, and the corresponding independent variable has a value of 2, 2) solution equation -x 2 + 4x-3 = 0, x -1) (x-3) = 0

x1=1,x2=3

So the coordinates of the intersection of the image and the x-axis of the function are (1,0),(3,0).

Recipe this function to get -(x 2-4x)-3

(x-2)^2+1

So this function has a minimum value of 1

Substituting y=0 into the analytic formula of the function The coordinates of the intersection of x=3 or x=1 are obtained: (1,0); （3，0）

Solution: (1).

y=-x²+4x-3

(x²-4x+4)+1

(x-2)²+1

x-2)²≥0

(x-2)²≤0

That is, when (x-2) = 0, that is, when the independent variable x=2, the maximum value of the function is 1, and there is no minimum value y=-x +4x-3=0

x²-4x+3=0

x-1)(x-3)=0

x-1 = 0 or x-3 = 0

x = 1 or x = 3

That is, the coordinates of the intersection point with the x-axis are (1,0), and the vertex coordinates of (3,0)(2)y=1 3(x+5) -3 are (-5,-3).

Vertex y=-(x-2) 2+1

The opening is downward, there is a maximum value of 1, at which point x=2, and the intersection points with the x-axis are (1,0), (3,0).

The second question can be written directly.

The vertex coordinates are (-5, -3).

y=-x2+4x-3

(x^2-4x)-3

(x^2-4x+4)+4-3

(x-2)^2+1

When x = 2, the maximum value of the function is 1

x^2+4x-3=0

x^2-4x+3=0

x-1)(x-3)=0

x1=1,x2=3

The coordinates of the intersection of the image and the x-axis of this function are (1,0), and the vertex coordinates of the parabola y=1 3(x+5) -3 are (-5,-3).

The maximum value is obtained at a fixed point:

Maximum: (4ac-b 2) (4a)=(4*3-16) (-4)=1 The value of the corresponding independent variable: x=-b (2a)=4 2=2x 2+4x-3=0

x^2-4x+3=0

x-3)(x-1)=0

x=3 or 1 intersection coordinates: (1,0); （3，0）

(1)y=(x-1)^2-4

*a is a coefficient of x 2 m is determined by the coefficient of x.

The opening is upward. **Determined by a factor of x2: a coefficient of x 2 with a coefficient of 0 with an opening up x 2 with a coefficient of 0 with an opening down.

Vertex coordinates (1,-4).

Axis of symmetry x=1

2) Let's draw it yourself.

*ps: The image intersects with the x-axis at (-1,0) and (3,0)(3) from the image as we can see that y has a minimum value of -4

4) From the image, we can see that y 0x= -1 when x -1 or x 3 or y=0 when x = 3

y 0 at 1 x 3

Solution:

y=-x²+4x-3=-(x²-4x+4)+1=-(x-2)²+1∵(x-2)²≥0∴-(x-2)²≤0That is, when (x-2) = 0, that is, the independent variable x=2, the maximum value of the function is 1, and there is no minimum value

y=-x²+4x-3=0x²-4x+3=0(x-1)(x-3)=0x-1 = 0 or x-3 = 0x = 1 or x = 3That is, the coordinates of the intersection point with the x-axis are (1,0),(3,0).

(2) y=1 3(x+5) The vertex coordinates of -3 are (-5, -3).

y=-x2+4x-3

(x^2-4x)-3

(x^2-4x+4)+4-3

(x-2)^2+1

When x = 2, the maximum value of the function is 1

x^2+4x-3=0

x^2-4x+3=0

x-1)(x-3)=0

x1=1,x2=3

The coordinates of the intersection of the image and the x-axis of this function are (1,0), and the vertex coordinates of the parabola y=1 3(x+5) -3 are (-5,-3).

Vertex y=-(x-2) 2+1

The opening is downward, there is a maximum value of 1, at which point x=2, and the intersection points with the x-axis are (1,0), (3,0).

The second question can be written directly.

The vertex coordinates are (-5, -3).

y=-(x-1)(x-3)

obtain: b(0,-3), a(1,0),c(3,0)ab= (1+3 2)= 10

ac=3-1=2

bc=√(3^2+3^2)=3√2

abc perimeter = ab + ac + bc = 2 + 10 + 3 2 abc area = 1 2 * ac*h = 1 2 * 2 * 3 = 3 can solve your problem?

Solution: The image of the quadratic function y=x 2-2x-3 is shown in the following figure:

1) The root of the quadratic equation x 2-2x-3=0 is the abscissa of the intersection point of the image and the x-axis of the quadratic function y=x 2-2x-3, and the root of the equation is x1=-1, x2=3, as shown in the image.

2) When x -2 or x = 4, the value of the function y is 5

Study hard and make progress every day. It's not a difficult question.