We know that the equation in the form of ax2+bx+c=0 is a quadratic equation, while the equation in the form of y= ax2+bx+c(a, B and C are constants, and a≠0) is a quadratic function. They are almost the same in form, the only difference is that the expression of quadratic equation in one variable is equal to 0, while the expression of quadratic function is equal to y, which makes the relationship between them particularly close, and many questions are based on this. Why is this happening? Mainly because when the variable y in the quadratic function takes 0, the quadratic function becomes a quadratic equation with one variable. It can be seen that many knowledge points in the equation can be applied to the function. Below, let's learn more about the specific applications between them.
First, the application relationship between solving equation and quadratic function by matching method.
One of the four ways to solve the equation is to solve the equation by matching method. In quadratic function, we often need to transform the general form into a new form. This transformation process is actually to formulate it, just like the equation formula.
Example 1: Solving Equation by Matching Method
Solution:
( 1)
(2)
(3)
(4)
……
Example 2: Point out the vertex coordinates of the function.
Solution:
(5)
(6)
(7)
(8)
∴ Vertex is (-2,-17)
The four steps (1), (2), (3) and (4) in the equation are exactly the same as those in the functions (5), (6), (7) and (8). It can be seen that the equation is closely related to the function.
We know from the study of textbooks; When the image of the quadratic function y= ax2+bx+c(a≠0) intersects the X axis, the value of the abscissa of the intersection point is the root of the equation ax2+bx+c=0(a≠0).
Second, the discriminant of the roots of quadratic equation in one variable and the application of quadratic function.
In quadratic function, when the function has two intersections, one intersection and no intersection with the X axis, the discriminant of the root of the quadratic equation of one variable corresponding to the function is: △ >; 0, delta = 0 and delta; 0, the equation has two unequal real roots; When △ = 0, the equation has two equal real roots; When △ < 0, the equation has no real root.
Example 3: Judge the number of quadratic function y = x2-4x+3 intersecting with the X axis.
Analysis: Because the number of intersection points of quadratic function and X axis can be determined by the discriminant △ of the root of the corresponding equation. If △ > 0, there are two intersections; If △ = 0, there is an intersection; If △
Example 4: Try to explain the function y = x2-4x+5, no matter what value X takes, y >；; 0。
Analysis: The first method is to use the matching method to explain in the form of Y = (x-2) 2+ 1 (But if the coefficient value is not good, this method will be more troublesome. )
The second method: use delta to explain, because delta =-4; 0, so the image opening is upward. So the image is on the X axis, so no matter what value X takes, y >；; 0。
Example 5: Prove that no matter what real number M takes, the equation x2-(m2+m) x+m-2 = 0 must have two unequal real number roots.
Analysis: if this problem is done in the conventional way, it is to prove the delta of a quadratic equation with one variable. The problem of 0. However, the discriminant △ of this problem is a univariate quartic polynomial about m, and the sign is difficult to judge, which brings trouble to the proof. If we analyze the meaning of the problem with the idea of function, let f (x) = x2-(m2+m) x+m-2, because its opening is upward, we only need to find a real number x0, so f (x0).
Pay attention to observation, it is easy to find that when x = 1, f (1) =1-(m2+m)+m-2 =-m2-1
This shows that the conclusion to be proved is valid.
Simply prove it.
Third, the application of the relationship between root and coefficient in the quadratic equation of one variable in function.
Example 6: Quadratic function images intersect (-1, 0) and (3,0), and the Y axis intersects at (0,3), so as to find the resolution function.
Analysis: The conventional solution to this kind of problem is the undetermined coefficient method. But it can be solved by the relationship between roots and coefficients, because (-1, 0) and (3,0) are actually on the X axis, so-1 and 3 are the two roots of the equation corresponding to the function.
Solution: Let the function form become
Function intersection (0, 3)
∴ c=3
∴
There are also: function intersections (-1, 0), (3, 0).
That is to say, the abscissas of the intersection of the function and the x axis are-1 and 3.
∴
The solution is a =- 1 and b = 2.
The function form is y =-x2+29x+3.
Obviously, this method is simpler than the undetermined coefficient method.
The close relationship between quadratic equation and quadratic function has many ingenious uses. There is only so much to discuss here, and more places need to be slowly realized in practice.
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