Make the following function image y = absolute value x square - 2x-1

Make the following function image y = absolute value x square - 2x-1

First draw the parabola y = x & # - 2x-1, and then fold the image below the x-axis to the image above the x-axis
What is the image of 2 / 2 of the absolute value of function y = x? Why
The image of the function y = 2 / 2 of the absolute value of X is shown in the following figure:
When x is positive, y is positive,
When x is negative, y is positive, so the image is in the first and second quadrants
Given the function f (x) = x2 + | x-a | + 1, a ∈ R. (1) try to judge the parity of F (x); (2) if - 12 ≤ a ≤ 12, find the minimum value of F (x)
(1) When a = 0, f (- x) = (- x) 2 + | - x | + 1 = f (x), then f (x) is even function. When a ≠ 0, f (a) = A2 + 1, f (- a) = A2 + 2 | + 1, f (a) ≠ f (- a), f (a) ≠ - f (- a), then f (x) is non odd and non even function
Judgment of parity of function with absolute value
If f (x) = | x-a | A is a real number, X belongs to R, judge its parity?
No matter what the expression is! Judging the parity of a function is in accordance with the following two steps: 1, see if the domain is symmetric about the origin! Asymmetry is no parity to say! If symmetric, do the next step! 2, judge the relationship between F (- x) and f (x), f (- x) = f (x) is an even function; f (- x) = - f (x) is an odd function
① What is the square of the difference between root 5 and root 2?
② The sum of root 6 divided by one-half of root and one-third of root?
③ : the 2007 power of the difference of 2 minus 3 multiplied by the 2009 power of the sum of 2 plus 3?
Speed I want it right now! Urgent
There's another question to ask, whether it's more or less? Is it the same? It's the second one above!
One
(2√5-√2)^2=(2√5)^2-4√10+2=22-44√10
Two
√6/(1/√2+1/√3)
=√6/[(√3+√2)/√6]
=√6/(√3+√2)
=√6(√3-√2)
Three
(2√2-3)^2007·(2√2+3)^2009
(2√2-3)·(2√2+3)=8-9=-1
So the original form
=(-1)^2007·(2√2+3)^2
=-(17+12√2)
If two groups of numbers x1, X2 , xn and Y1, Y2 The average of them is. X and. Y respectively, then a new group of numbers X1 + Y1, X2 + Y2 The average number is n + xyn______ .
In this paper, the author gives a brief introduction to the meaning of the title xn）=.x，1n（y1+y2+… A new set of numbers X1 + Y1, X2 + Y2 The average of XN + yn = 1n (x1 + Y1 + x2 + Y2 +...) +xn+yn）=1n（x1+x2+… xn）+1n（y1+y2+… Yn) =. X +. Y. so the answer is. X +. Y
It is known that the vertex coordinates of triangle ABC are a (4, - 6) B (- 4,0) C (- 1,4) respectively. The equation for finding the line where the median line PQ of triangle ABC parallel to ab side lies is obtained
The abscissa of the midpoint of AC is 4 + (- 1) / 2 = 5 / 2, and the ordinate is - 6 + 4 / 2 = - 1
The abscissa of the midpoint of BC is - 4-1 / 2 = - 5 / 2, and the ordinate is 0 + 4 / 2 = 2
So the median line goes through (5 / 2, - 1), (- 5 / 2,2)
The linear equation is y = - 8 / 5x + 1
Find the limit of sequence Lim = [(an ^ 2 + bn-1) / (4N ^ 2-5n + 1)] = 1 / B and find the value of a and B
1lim = [(an ^ 2 + bn-1) / (4N ^ 2-5n + 1)] = 1 / B to find the value of a and B
2lim = [1 / (A-1) ^ 2] = 0 to find the value range of a
3lim [1 + (a + 1) ^ n] = 1 the range of a
N tends to infinity
Click enlarge, then click enlarge again:
What is the square of the difference between root 2 and 3?
The answer is 1
How to prove "| x + X1 + x2 + ·· + xn | ≥ | x · - (| x1 · + | x2 · + ·· + | xn ·)"?
It can be obtained from trigonometric inequality|
x+x1+x2+···+xn|≥|x|- |x1+x2+···+xn|
Because | X1 + x2 + ··· + xn|
Using trigonometric inequality:
lx±yl≤lxl+lyl
First get:
|x+x1+x2+···+xn|+lx1+x2+···+xnl≥|xl
Combined with | x1 | + | x2 | + ·· + | XNL ≥ LX1 + x2 + ·· + XNL
You can prove the conclusion.