The image of X-2 power + 1 (a > 0, a is not equal to 1) of function y = a must pass through the point?

The image of X-2 power + 1 (a > 0, a is not equal to 1) of function y = a must pass through the point?

y=a^(x-2)+1
When x = 2, y = a ^ 0 + 1 = 1 + 1 = 2
So we have to go through (2,2)
Is the distance from a fixed point on an ellipse to two focal points constant, equal to the long half axis?
This is the first definition of an ellipse
Pf1 + PF2 = 2A, F 1 and F 2 are the focal points, P is the point on the ellipse, and a is the length of the semimajor semiaxis
yes
If a > 0, a is not equal to 1, then the image of function y = a ^ (x-1) + 1 must pass through the point
A is an unknown number whose value affects the value of Y
Substituting x makes the formula not appear unknowns and get the fixed value
When x = 1, y = 2
Fixed point (1,2)
Idea: the original point of exponential function is (0, 1)
Move the point one to the right and one up
Then the final point is (1,2)
Given the elliptic equation x ^ 2 / 4 + y ^ 2 = 1, if the distance between the point P on the ellipse and the left focus is 3, then the distance between the point P and the right focus is 3
Elliptic equation x ^ 2 / 4 + y ^ 2 = 1
a²=4,a=2
Let the left and right focus be f1.f2 respectively
Then | Pf1 | = 3
∵ P on an ellipse
∴|PF1|+|PF2|=2a=4
∴|PF2|=4-|PF1|=4-3=1
That is, the distance from point P to the right focus is 1
If a > 0 and a is not equal to 1, then the image of function y = a ^ x + 3 - 4 must pass the point ()
Exponential function y = a ^ x constant crossing point (0,1)
｜ y = a ^ (x + 3) constant crossing point (- 3,1)
｜ y = a ^ (x + 3) - 4 constant crossing point (- 3, - 3)
(-3，-4）
The function is a straight line, so it must intersect the axis. When x = 0, it intersects at (0, - 1); when y = 0, it intersects at (1 / A, 0). Because a is uncertain, the image of the function must be (0, - 1)
It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 (a > b > 0) is (√ 6) / 3, and the distance from one end point of the minor axis to the right focus is √ 3
1. Find the equation of ellipse C
2. Let the line L and the ellipse C intersect at two points a and B, and the distance from the coordinate origin o to the line is √ 3 / 2, then the maximum area of the triangle AOB can be obtained
|Ab | = {(1 + K ^ 2)} [(x1 + x2) ^ 2-4x1x2] = {(1 + K ^ 2)} [36K ^ 2B ^ 2 / (1 + 3K ^ 2) ^ 2-4 · 3 (b ^ 2-1) / (1 + 3K ^ 2)] = {(1 + K ^ 2)} [(36K ^ 2 + 12-12b ^ 2) / (1 + 3K ^ 2) ^ 2] substitute B ^ 2 = 3 (k ^ 2 + 1) / 4 into = {(1 + K ^ 2)} [(36K ^ 2 + 12-9 (K ^ 2 + 1)) / (1 + 3K ^ 2) ^ 2] = {}}
If the function f (x) = a Λ (x-1) (x ≥ 0) is known, the image passes through the point (2,1 / 2), where a > 0 and a is not equal to 1
Is it the value of a
Substituting (2,1 / 2) into a Λ (x-1)
a^1=1/2
a=1/2
It is known that the eccentricity of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is (√ 3) / 2, and the distance from one end point of the minor axis to the right focus is 2
The equation of finding ellipse
If P (x, y) is a moving point on the ellipse, find the value range of 3x-2
I'd like to ask the questioner if he has copied the question wrong and if he wants to find the value range of 3x-2y. Otherwise, the question is too simple
First of all, read the title, and you will know immediately that the content of the title is to give you an elliptic equation, which only contains unknown quantities of a and B
It can be listed by "eccentricity is (√ 3) / 2": C / a = (√ 3) / 2
It can be listed by "the distance from one end point of the minor axis to the right focus is 2": a = 2
From ABC three relations: A ^ 2 = B ^ 2 + C ^ 2
B = 1
So the standard equation of ellipse is: x ^ 2 / 4 + y ^ 2 = 1
Therefore, any point on the ellipse can be set as: P (2cost, Sint)
So the original problem can be transformed into finding the value range of 6cost-2sint
And 6cost-2sint ≤ √ (6 ^ 2 + 2 ^ 2) = 2 √ 10
So the range of 3x-2y is [- 2 √ 10,2 √ 10]
Given the function F X = a ^ X / b (a > 0, a is not equal to 1), the image passes through a (4,1 / 2) B (5,1)
Finding the analytic expression of function FX
It is proved that the function y = log2 [1-f (x)] decreases monotonically in its domain of definition
（1） f(4)=a^4/b=1/2
f(5)=a^5/b=2
f(5)/f(4)=a=4
So B = 2.4 ^ 4 = 512
So f (x) = 4 ^ X / 512 = 2 ^ (2x-9)
(2) Let 1-f (x) > 0, that is 2 ^ (2x-9)
1/2=a^4/b .....(1)
1=a^5/b .....(2)
(2)/(1): 2=a, b=2^5=32
f(x)=2^x/2^5=2^(x-5)
y=log2[1-f(x)]=log2[1-2^(x-5)]
y'=[1-2^(x-5)]'/([1-2^(x-5)]ln2)
=-2^(x-5)ln2/([1-2^(x-5)]ln2)
=-2^(x-5)/[1-2^(x-5)]
By definition, 1-2 ^ (X-5) > 0, 2 ^ (X-5) > 0
So y '0, that is 2 ^ (2 × - 9)
One vertex of the ellipse is (0,2), and the distance between the right focus (C, 0) and (root 2, root 2) is 2
If the right focus is (C, 0), then a vertex of the focus on the x-axis is (0,2), then B = 2, B & # 178; = 4. The distance between the right focus (C, 0) and (√ 2, √ 2) d = √ [(C - √ 2) &# 178; + (0 - √ 2) &# 178;] = 2 = > √ (C & # 178; - 2 √ 2C + 2 + 2) = 2 = > C & # 178; - 2 √ 2C + 4 = 2 = > C & # 178; - 2 √ 2C + 2 = 0 = > C = √
From the distance formula between two points, we get 2 = √ (C - √ 2) ^ 2 + (0 - √ 2) ^ 2, and the solution is C = 2 √ 2
So we can get the right focus of the ellipse (2 √ 2,0) and the coordinate system is established with the X axis as the major axis
So the short half axis is b = 2
So B ^ 2 = 4, C ^ 2 = 8 = > A ^ 2 = 12
So the elliptic equation is x ^ 2 / 12 + y ^ 2 / 4 = 1
Let the elliptic equation be X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1, (a ＞ 0, B ＞ 0) because the title says that it is the right focus, so the ellipse has two focuses on the X axis
So it is a > b, a & # 178; = B & # 178; + C & # 178;, because a vertex is (0,2), so B = 2,
From the meaning of the question, we can get (C - √ 2) &# 178; + (0 - √ 2) &# 178; = 4, and get C = 0 (rounding off) or C = 2 √ 2
So a & # 178; = B &... Expand
Let the elliptic equation be X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1, (a ＞ 0, B ＞ 0) because the title says that it is the right focus, so the ellipse has two focuses on the X axis
So it is a > b, a & # 178; = B & # 178; + C & # 178;, because a vertex is (0,2), so B = 2,
From the meaning of the question, we can get (C - √ 2) &# 178; + (0 - √ 2) &# 178; = 4, and get C = 0 (rounding off) or C = 2 √ 2
So a & # 178; = B & # 178; + C & # 178; = 4 + 8 = 12
So the elliptic equation is X & # 178 / 12 + Y & # 178 / 4 = 1
If it's not detailed or wrong, you can continue to ask