"Quadratic functions and quadratic equations and inequalities" Quadratic functions, equations and inequalities PPT

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"Quadratic functions and quadratic equations and inequalities" Quadratic functions, equations and inequalities PPT

Part One: Explanation of Curriculum Standards

1. Understand the practical significance of the quadratic inequality of one variable.

2. Be able to solve quadratic inequalities of one variable with the help of quadratic functions of one variable; and be able to use sets to represent the solution sets of quadratic inequalities of one variable.

3. With the help of images of quadratic functions of one variable, understand the connection between quadratic inequalities of one variable and corresponding functions and equations.

Quadratic functions and quadratic equations and inequalities PPT, part 2: independent preview

1. The concept of quadratic inequality of one variable

1. Judging from the number of unknowns and the highest degree of the unknowns, what do the inequalities x2-2x-3>0, x2+5x≤0, -3x2-6x+1<0, 4x2-1≥0, etc. have in common?

Tip: They contain only one unknown, and the highest degree of unknown is 2.

2. Fill in the blanks

The concept and form of quadratic inequality of one variable

(1) Concept: We call an inequality that contains only one unknown number and the highest degree of the unknown number is 2, called a quadratic inequality of one variable.

(2)Form:

①ax2+bx+c>0(a≠0);

②ax2+bx+c≥0(a≠0);

③ax2+bx+c<0(a≠0);

④ax2+bx+c≤0(a≠0).

(3) Solution set: Generally, the value of x that makes a quadratic inequality of one variable true is called the solution of this inequality, and the set of all solutions of the quadratic inequality of one variable is called the solution set of the quadratic inequality of one variable.

3. Do it

The following inequalities are known: ①ax2+2x+1>0; ②x2-y>0; ③-x2-3x<0; ④ >0. The number of quadratic inequalities of one variable is ()

A.1 B.2 C.3 D.4

Analysis: When a=0 in ①, it is not a quadratic inequality of one variable; there are two unknowns in ②, it is not a quadratic inequality of one variable; ③ is a quadratic inequality of one variable; ④ is a fractional inequality.

Answer:A

2. Solution to quadratic inequality of one variable

1. (1) What is the zero point of the quadratic function y=ax2+bx+c? Is the zero point a point?

Tip: The real number x that makes ax2+bx+c=0 is called the zero point of the quadratic function y=ax2+bx+c. The zero point is not a point, but a real number. The zero point is the root of the corresponding equation of the function.

(2) The graph of the quadratic function y=x2-5x is as shown in the figure.

When x has what value, y=0? When x has what value, y<0? When x has what value, y>0.

What is the relationship between the function graph and the x-axis in the above situations?

Tip: When x=0 or x=5, y=0. At this time, the image and the x-axis intersect at two points (0,0) and (5,0);

When 0

When x<0 or x>5, y>0. At this time, the function graph is above the x-axis, and at this time x2-5x>0.

(3) For any quadratic inequality of one variable, what are the key points of the solution set?

Tips: ① The position of the parabola y=ax2+bx+c and the x-axis is the root of the quadratic equation ax2+bx+c=0; ② The opening direction of the parabola y=ax2+bx+c is also It is the positive and negative of a.

(4) What are the relative positions of the parabola y=ax2+bx+c (a>0) and the x-axis? How to use a quadratic equation to explain these positional relationships?

Tip: The parabola y=ax2+bx+c (a>0) and the x-axis may have two intersection points (intersection), one intersection point (tangent), and no intersection point (separation). It can be distinguished by the corresponding quadratic equation of one variable Judgment is based on the relationship between formula Δ and 0.

Quadratic functions and quadratic equations of one variable and inequalities PPT, the third part: exploration and learning

Solution of Quadratic Inequality of One Variable

Example 1 Solve the following inequalities:

(1)2x2-3x-2>0;

(2)-3x2+6x-2>0;

(3)4x2-4x+1≤0;

(4)x2-2x+2>0.

Analysis: First find the solution to the corresponding quadratic equation of one variable, and then combine the image of the corresponding quadratic function to write the solution set of the inequality.

Solution: (1) The solution to the equation 2x2-3x-2=0 is x1=-1/2, x2=2.

Because the graph of the corresponding function is a parabola that opens upward,

So the solution set of the original inequality is {x├|x<"-" 1/2 "or " x>2}┤.

(2) The inequality can be reduced to 3x2-6x+2<0.

Because the discriminant of 3x2-6x+2=0 Δ=36-4×3×2=12>0, the solution of equation 3x2-6x+2=0 is x1=1-√3/3,x2=1+ √3/3.

Because the graph of the function y=3x2-6x+2 is a parabola that opens upward, the solution set of the original inequality is {x├|1"-" √3/3

(3) The solution to the equation 4x2-4x+1=0 is x1=x2=1/2. The graph of the function y=4x2-4x+1 is a parabola that opens upward, so the solution set of the original inequality is {x├& #124;x=1/2}┤.

(4) Because the discriminant Δ<0 of x2-2x+2=0, the equation x2-2x+2=0 has no solution. And because the graph of the function y=x2-2x+2 is a parabola that opens upward, so The solution set of the original inequality is R.

Reflection on general steps for solving quadratic inequalities of one variable without parameters

(1) Standardization. By deforming the inequality, make the right side of the inequality 0 and make the coefficient of the quadratic term positive.

(2) Discriminant. Factor the left side of the inequality. If it cannot be factored, calculate the discriminant of the corresponding equation.

(3) Find the real roots. Find the roots of the corresponding quadratic equation or show that the equation has no real roots based on the discriminant.

(4) Draw a sketch. Draw a sketch of the corresponding quadratic function based on the roots of the quadratic equation of one variable.

(5) Write the solution set. Write the solution set of the inequality based on the image.

Quadratic functions and quadratic equations of one variable and inequalities PPT, the fourth part: thinking analysis

Common methods for finding parameter ranges in problems where inequalities are constant

1. Use the discriminant of the roots of a quadratic equation to solve the problem of constant establishment of a quadratic inequality on R.

Suppose f(x)=ax2+bx+c(a≠0), then

f(x)>0 is always established⇔{■(a>0"," @Δ<0"," )┤f(x)≥0 is always established⇔{■(a>0"," @Δ≤0" ," )┤

f(x)<0 is always true⇔{■(a<0"," @Δ<0"," )┤f(x)≤0 is always true⇔{■(a<0"," @Δ≤0" ." )┤

When the unspecified inequality is a quadratic inequality of one variable, there is

(1) The inequality ax2+bx+c>0 is always true for any real number x ⇔{■(a=b=0"," @c>0)┤ or {■(a>0"," @Δ<0" ;" )┤

(2) The inequality ax2+bx+c<0 is always true for any real number x⇔{■(a=b=0"," @c<0)┤or {■(a<0"," @Δ<0" ." )┤

2. Separate the independent variables and parameter variables, and use the idea of ​​​​equivalence transformation to transform the original problem into the problem of finding the optimal value of the function.

Quadratic functions and quadratic equations and inequalities PPT, part 5: in-class exercises

1. The solution set of the inequality x2-9<0 is ()

A.{x|x<-3} B.{x|x<3}

C.{x|x<-3 or x>3} D.{x|-3

Analysis: From x2-9<0, we can get x2<9, and the solution is -3

Answer:D

2. If the solution set of the inequality 4x2+ax+4>0 is R, then the value range of the real number a is ()

A.(-16,0) B.(-16,0]

C.(-∞,0) D.(-8,8)

Analysis: The solution set of inequality 4x2+ax+4>0 is R, ∴Δ=a2-4×4×4<0, the solution is -8

Answer:D

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Update Time: 2024-07-03

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