What properties can be summarized from the images of quadratic functions y = x ^ 2 and y = - x ^ 2?

What properties can be summarized from the images of quadratic functions y = x ^ 2 and y = - x ^ 2?

The opening of a & gt; 0 parabola is upward, the opening of a & lt; 0 parabola is downward; the symmetry axes are Y-axes, and the vertex coordinates are (0,0). These two function images are symmetric about x-axis and are centrosymmetric,

Why is y = a (X-H) ^ 2 + k of quadratic function the vertex form? How was it obtained at the beginning

Because in (h, K), the highest or lowest point of the function is it

In the exam, the analytic formula of quadratic function is y = a (X-H) &# 178; + K
Can it be directly used as the final answer of the analytic formula, and be reduced to y = ax & # 178; + BX + C?

Generally speaking, our teacher said at that time that the vertex formula y = a (X-H) ² + K could not be changed into the general formula y = ax & #178; + BX + C
But the intersection formula y = a (x-x1) (x-x2) must be changed into a general formula

Why does the quadratic coefficient a determine the opening direction and size of the parabola

Please draw the following function image on standard checkerboard paper to make sure you understand immediately
y=x²
y=-x²
y=2x²
y=-2x²
y=x²+2x+3
y=2x²+1
You will find that when | a | is the same, the size after opening is the same, when the symbol of a is the same, the opening direction is the same, the positive number opening up, the negative number opening down
So a determines the direction and size of the opening

It is known that the point P is a point in the second quadrant of the image with the inverse scale function y = K / X. through the point P, the vertical line of the x-axis and the y-axis is made. The vertical feet are m, n respectively,
If the area of rectangular ompn is 5, then K is calculated?

∵ the image of inverse scale function y = K / X is in the second quadrant
∴k﹤0
∵ point P is the point of inverse scale function y = K / X in the second quadrant, and the area of rectangular ompn is 5
∴|k|=5
∴k=-5

It is known that the circle a (x-3) 2 + (Y-3) 2 = 4, passing through the point P as its tangent intersects at the point m, so that PM = Po. When the value of P is, there is the shortest PM and the shortest PM length is

If O is a dot (0,0)
Let the coordinates of point p be (x, y)
Then Po ^ 2 = x ^ 2 + y ^ 2
pm^2=(x-3)^2+(y-3)^2-R^2
pm=mo
Then x ^ 2 + y ^ 2 = (x-3) ^ 2 + (Y-3) ^ 2-4
3x + 3Y = 1 is the locus of point P
PM is the shortest, that is, Po is the shortest
The shortest distance is the distance from O to the line 3x + 3Y = 1
It is easy to get that P is (1 / 6,1 / 6) and the distance is √ 2 / 6

It is known that the circle a (x-3) 2 + (Y-3) 2 = 4, passing through the point P as its tangent intersects at the point m, so that PM = Po. When the value of P is, there is the shortest PM and the shortest PM length

Let P (x, y) be PM = Po
Get x ^ 2 + y ^ 2 = (x-3) ^ 2 + (Y-3) ^ 2-4 (^ is a power sign)
We can get l: x + y = 7 / 3
The shortest PM is the shortest Po
Then the problem is to find the distance from O to L
It can be seen that PM = 7 / 6 * root 2

If the circumference of the triangle AOB is 12, the equation of the line is solved

Let the coordinates of a and B be a (a, 0) and B (0, b), respectively. Then the perimeter of triangle AOB is s = a + B + √ (A & # 178; + B & # 178;) = 12. Let the linear equation be y = - (b / a) x + B. If a straight line passes (4 / 3,2), then 2 = - (B / a) 4 / 3 + B, 6a + 4B = 3AB, and a + B + √ (A & # 178; + B & # 178;) = 12. Then a and B can be solved by substituting into the equation

To solve the equation, if the difference between 18 / 3 + m and m-2 / 5 of the algebraic formula is 8, find the value of M

The difference between 18 / 3 + m and m-2 / 5 is 8
90+5m-(3m-6)=120
2m=24
m=12

In the plane rectangular coordinate system xoy, the image of the first-order function y = KX + B passes through points (0,2) and intersects with the positive half axis of X axis at points a, P and Q on the line segment Ao,
The triangle OPM and the triangle QMN are two isosceles right triangles with the similarity ratio of 3:1. The angle OPM = angle mqn = 90 is used to find the value of an: am and the image expression of the first-order function y = KX + B

The results are as follows: (1) ∵ PM ∥ QN, ∵ an ∶ am = QN ∶ PM = 1 ∶ 3, that is am = 3an
(2) The coordinates of ∵ B are (0,2), ∵ B = 2
Mn = am-an = 3an-an = 2An, OM = 3MN,
So om = 3MN = 6An
OA=0M+MN+AN=6AN+2AN+AN=9AN
Let y = KX + 2 = 0, then x = OA = - 2 / K
So 9An = - 2 / K
That is an = - 2 / 9K
Let the coordinates of point p be (XP, YP), p be on AB, so YP = kxp + 2, and p be on Op,
So YP = XP, that is, kxp + 2 = XP, XP = - 2 / (k-1). (2)
XP=OM/2=6AN/2=3AN=3(-2/9K)=-2/3K.(3)
From (2) (3) - 2 / 3K = - 2 / (k-1)
3k=k-1,2k=-1,∴k=-1/2.
So the first-order function is y = (- 1 / 2) x + 2