If we know that a and B are on the image of quadratic function y = AX2, the abscissa of these two points are - 2, 1 respectively, and the triangle AOB is a right triangle (o is the origin of coordinates), we can find the value of A

If we know that a and B are on the image of quadratic function y = AX2, the abscissa of these two points are - 2, 1 respectively, and the triangle AOB is a right triangle (o is the origin of coordinates), we can find the value of A

Substituting the abscissa of a and B into the analytical formula, the ordinate is obtained
A is (- 2,4a) B is (1, a)
OA, OB cross the origin (0,0),
OA slope K1 = (4a-0) / (- 2-0) = - 2A
OB slope K2 = (A-0) / (1-0) = a
OA ⊥ ob
So K1 * K2 = - 1
-2a*a=-1
A = ± root 2 / 2
Y is a standard parabola, and the vertex is the origin (0,0).
AB is on the parabola, substituting its abscissa to y, the ordinate of AB is 4a, a
Let's find the coordinates o (0,0), a (- 2,4a), B (1, a) of three points of a triangle, and let the triangle be a right triangle
There are many ways to solve it. For example, the same triangle can be solved by the method of equal area,
The distance from the origin o to AB is high, AB is the bottom; OA and ob are high and bottom; anyway, only one a is unknown
Y is a standard parabola, and the vertex is the origin (0,0).
AB is on the parabola, substituting its abscissa to y, the ordinate of AB is 4a, a
Let's find the coordinates o (0,0), a (- 2,4a), B (1, a) of three points of a triangle, and let the triangle be a right triangle
There are many ways to solve it. For example, the same triangle can be solved by the method of equal area,
The distance from the origin o to AB is high, AB is the bottom; OA and ob are high and bottom; anyway, only one a is unknown
Basically, from the topic, you can determine that the triangle is a triangle with angle o as the right angle, and then substitute the coordinates of a and B. there should be a lot of methods to follow, such as slope method, area method, Pythagorean theorem method, cosine function method, vector method, etc
Two hundred and fifty-two
If the image of x power + 1 of function y = 2 is translated into the image of x power + 1 of function y = 2 according to vector a, then a=
Let a = (m, n)
After translating the image of function y = 2 ^ x + 1 according to vector a, the image of function y = 2 ^ (x-m) + 1 + n is obtained,
Compared with y = 2 ^ (x + 1), M = - 1, n = - 1
∴a=(-1,-1).
X → x + 1 shift left by 1
Y-1 → y down 1
A = (- 1, - 1) question: I know that left plus right minus, up plus down minus, but I want to ask why x → x + 1 moves left by 1, why not (1, - 1), why left plus is not + 1, thank you.
It is known that the abscissa of two points a and B on the image of quadratic function y = a ^ 2 (a is greater than or equal to 0) are - 1 and 2 respectively, and point O is the origin of the coordinates,
If the triangle AOB is a right triangle, the perimeter of the triangle AOB is
1) Function can be set to
y=k(x-1)(x-3)
Because the function passes the point (0,3 / 2)
So k = 1 / 2
So the expression of quadratic function is
y=(x-1)(x-3)/2
=(x^2-4x+3)/2
=x^2/2-2x+3/2
(2) Counter evidence
Let there be a real number m and a point m (m, - m ^ 2) on the graph of this quadratic function
Then - m ^ 2 = m ^ 2 / 2-2m + 3 / 2
3m^2/2-2m+3/2=0
(3m^2-4m+3)/2=0
3m^2-4m+3=0
At this time, we can know that delta = - 20
A(-1,a) B(2,4a)
Because it's vertical, so - A * 2A = - 1,
A = √ 2 / 2
OA = √ 6 / 2
OB=2*(√3) AB=3*(√6)/2
Perimeter = 2 * (√ 6) + 2 (√ 3)
I wish you every day
The x-th power image of the function y = 4 is translated according to the vector a = (2, - 2) to obtain F, and the function analytic expression of F is obtained
The coordinate of vector a is regarded as two units to the right in X direction and two units to the down in Y direction
According to the principle of left plus right minus, because your function is an exponential function, I don't know much about it, but the general quadratic function is translated in this way~
The initial position of RT triangle ABC in the plane rectangular coordinate system is the angle ABC = 90 °, ab = 6, AC = 3, a, B, C, 3 points on the circle, AB is the diameter, G is the center, a
It coincides with the coordinate origin o, when a starts to slide to the right from the origin on the x-axis, and at the same time, point B also follows on the y-axis
When B coincides with the origin o, the movement ends. In the above movement, circle g always takes AB as the diameter. Question 1 the position relationship between original O and circle g. set the coordinate of point C as [x, y]. Find the relationship between Y and X. write the path length of point C in the whole city movement
It should be that the hypotenuse is for the diameter. AB is the diameter. The three points can't be on the circle with u and G as the center
1, the circumference angle of diameter is 90
2, y = root 3 * x
3,3
Where is the soy sauce
How to translate the function y = 2x to y = (2x) / 2?
y=2^x/2=2^(x-1)
So just move y = 2 ^ x one unit to the right
How to translate the function y = 2x (power) to y = (2x) / 2
It's impossible!!!
No way out,
The position of RT △ AOB in the plane rectangular coordinate system is shown in the figure. Point O is the origin, point a (0,8), point B (6,0), point P is on the line AB, and AP = 6. (1) calculate the coordinates of point p; (2) whether there is a point Q on the X axis, so that the triangle with B, P, Q as the vertex is similar to △ AOB. If there is, request the coordinates of point Q. if not, please explain the reason
(1) From Pythagorean theorem, we get AB = 10, let P point coordinate be (x, y), then apab = xob can be obtained from triangle similarity, and x = 3.6 can be obtained by substituting numerical value. Ab − apab = yoa, y = 3.2, so P point coordinate is (3.6, 3.2). (2) suppose Q point coordinate is (Q, 0), if BP is hypotenuse, then q = 3.6
What transformation can we get the 1-2x power of function y = 2 by transforming the image of function y = 4 to - x power
The - x power of y = 4 = (1 / 4) ^ x, y = 2 ^ (1-2x) = 2 * 2 ^ (- 2x) = 2 * (2 ^ (- 2) ^ x) = 2 * (1 / 4) ^ X. therefore, we can get the 1-2x power of the function y = 2 by changing the image of the - x power of the function y = 4 to make its abscissa unchanged and its ordinate to 2 times of the original
The vertices a and C of RT △ ABC move on the y-axis respectively (C does not coincide with the origin o), AC = 4 and BC = 2. Note that the angle between BC and the positive half axis of y-axis is a
(1) The length of ob is calculated for a = 90 ° and 45 ° respectively
(2) Some people say that no matter how a changes, the length of ob will never change. Is this correct? If it is, Party A and Party B will prove that if it is not, give a counter example
The condition is incomplete and the o-point is uncertain
The condition is incomplete and the o-point is uncertain
How to translate the image of function y = - 2x to the power of 2 to get the image of y = - 2x + 2x-3 / 2
How to translate the image of the function y = - 2x ^ 2 to get the image of y = - 2x + 2x-3 / 2
y=-2x^2+2x-3/2=-2(x^2-x)-3/2=-2(x^2-x+1/4-1/4)-3/2=-2(x-1/2)^2+1/2-3/2=-2(x-1/2)^2-1
So move y = - 2x ^ 2 1 / 2 to the right and 1 / 2 down