Given the exponential function f (x) = a ^ x (a > 0 and a is not equal to 1), the image passes through points (2,4), and then f (0) f (1) is obtained

Given the exponential function f (x) = a ^ x (a > 0 and a is not equal to 1), the image passes through points (2,4), and then f (0) f (1) is obtained

Bring the dots in
The analytic expression of the function is f (x) = 2 ^ X
Then f (0) = 1
F(1)=2

It is known that there are two symmetries about the line L: y = K (x-1) + 1 on the parabola y2 = x, and the value range of the real number k is obtained
I don't want to copy it. I see that the person who answered the previous question read the wrong question. When solving the question, I try to write it in detail, which is of great significance
I'm asking the range of K. you're wrong

When the line L: y = K (x-1) + 1K ≠ 0, let the line L ': y = - 1 / KX + my = - 1 / KX + m and Y & # 178; = x stand side by side, eliminate x to get: y = - 1 / KY & # 178; + m is Y & # 178; + KY km = 0, Δ = K & # 178; + 4km > 0, let L' intersect the parabola at a (x1, Y1), B (X2, Y2) a, B midpoint m (x0, Y0), a, B about l symmetry, then 2y0 = y

Factorization of X3 + X + 2 x3-4x2-4x-5

x3+x+2
=x^3+x^2-x^2+x+2
=x^2(x+1)-(x-2)(x+1)
=(x+1)(x^2-x+2)
x3-4x2-4x-5
=x^3-5x^2+x^2-4x-5
=x^2(x-5)+(x-5)(x+1)
=(x-5)(x^2+x+1)

1. List the calculation. (1) subtract the product of 10.5 and 0.8 from 11.4. What is the quotient of the difference divided by 1.5?
(2) What is the difference between 6.8 and 1.4 plus the quotient of 1.25 divided by 2.5?

(11.4-10.5x0.8)/1.5=2
6.8-1.4+2.5/1.25=7.4

The equation of the circle passing through the intersection of the line 2x-y + 1 = 0 and the circle x2 + y2-2x-15 = 0 and the origin is______ ．

Let the equation of the circle be x2 + y2-2x-15 + λ (2x-y + 1) = 0, because the circle passing through the intersection of the straight line 2x-y + 1 = 0 and the circle x2 + y2-2x-15 = 0 passes through the origin, so we can get - 15 + λ = 0, and the solution is λ = 15. By substituting λ = 15 into the equation and simplifying, we can get the equation of the circle: x2 + Y2 + 28x-15y = 0. So the answer is: x2 + Y2 + 28x-15y = 0

X ^ 2 + 4x-9 = 2x-11 to solve the equation

x^2+4x-9=2x-11
x^2+4x-2x-9+11=0
x^2+2x+1+1=0
(x+1)^2+1=0
There is no solution in the range of real number

Given that P: - 2 ≤ x ≤ 10; Q: x2-2x + 1-m2 ≤ 0 (M > 0), if P is a necessary and insufficient condition of Q, the value range of real number m is obtained

p: If P is a necessary and insufficient condition for Q, that is, "Q {P" ⇔ {x | 1-m ≤ x ≤ 1 + m} ⊊ {x | - 2 ≤ X - (1-m)) (x - (1 + m)) ≤ 0 ⇔ 1-m ≤ x ≤ 1 + m} ⊊ {x | - 2 ≤ x ≤ 10}, ∪ 1 − m ≥ − 21 + m ≤ 10, ∪ m ≤ 3, and M > 0, the value range of real number m is 0 < m ≤ 3

5 out of 13 times 14

5/13×14=5/13×（13+1）
=5/13×13+5/13×1
=5 and 5 / 13

It is known that the parabolic equation y = - & frac 12; X + H, points a, B, P (2,4) are all parabolic points, and the inclination angles of straight lines PA and Pb are complementary
(1) Verification: the slope of line AB is a fixed value
(2) When the longitudinal intercept of line AB is greater than zero, the maximum value of △ PAB area is obtained
If you answer right, you'll get bonus points and rewards

【1】 It is proved that: (1) on the parabola y = (- 1 / 2) x & sup2; + H, point P (2,4) is 4 = (- 1 / 2) × 2 & sup2; + H
∴h=6.
The parabola y = (- 1 / 2) x & sup2; + 6
② ∵ points a and B are on the parabola, so we can set their coordinates as a (2a, 6-2a & sup2;), B (2B, 6-2b & sup2;) (a ≠ b)
③ According to the problem, if the inclination angle of the straight line PA is β, then the inclination angle of the straight line Pb is π - β
The slope formula shows that kPa = Tan β. KPB = Tan (π - β) = - Tan β
That is to say, the sum of the slopes of the two lines PA and Pb is 0
From the slope formula, it can be obtained that kPa = (2-2a & sup2;) / (2a-2) = - (a + 1)
Kpb=(2-2b ²)/(2b-2)= －(b+1).
∴[-(a+1)]+[-(b+1)]=0.∴a+b=-2.
④ According to the slope formula, KAB = [(6-2a & sup2;) - (6-2b & sup2;)] / (2a-2b) = (B & sup2; - A & sup2;) / (a-b) = - (a + b) = 2
The slope of line AB is constant 2
① The slope of the straight line AB is 2, so we can set its "oblique equation" as y = 2x + t
Moreover, the longitudinal intercept of line AB is positive, t > 0
The simultaneous parabolic equation y = (- 1 / 2) x & sup2; + 6 and the linear equation y = 2x + t
x ²+4x+2(t-6)=0.
The discriminant ⊿ = 16-8 (T-6) = 8 (8-t) > 0. 0 < T < 8
② According to the "chord length formula of conic curve", chord | ab | = √ [40 (8-t)]
According to the formula of distance from point to line, the distance d from point P (2,4) to line AB: y = 2x + T is:
d=t/(√5).
The area of triangle ⊿ PAB s = (1 / 2) ×| ab | × d = (1 / 2) × √ [40 (8-t)] × T / (√ 5)
=√[2t²(8-t)]= √[2(-t³+8t²)].
③ Now let's find the maximum value of the function f (T) = - T & sup3; + 8t & sup2;, (0 < T < 8)
The derivation gives f ′ (T) = - 3T & sup2; + 16t. = - t (3T-16)
It is easy to know that in the interval (0,8), when 0 < T < 16 / 3, f ′ (T) > 0
When 16 / 3 < T < 8, f '(T) < 0
It can be seen from "the relationship between the monotonicity of function and the positive and negative of its derivative",
When t = 16 / 3, the area of ⊿ PAB is the largest
④ When t = 16 / 3, s = (64 √ 3) / 9 can be obtained from S = √ [2T & sup2; (8-t)]
The maximum value of ⊿ PAB area is (64 √ 3) / 9

A simple and convenient method to reduce 13 by 7 / 8 (8.16 by 1.875) by 0.84

A simple and convenient method to reduce 13 by 7 / 8 (8.16 by 1.875) by 0.84
=13 + 7/8 - 8.16 + 1+ 7/8 - 0.84
=13 + 1 - (8.16 + 0.84)+ 7/8+ 7/8
= 14 - 9 + 14/8
= 5 + 7/4
=Six and three fourths