Drawing function image with absolute value （1）y=|x-5|+|x+3| (2) Y = 2x-3 x ∈ Z | x | contained in Z （3）y=x²-2|x|-1 （4）y=｛x²+2x（x≥0） -x²-2x（x＜0） The best way to draw these four functions is to teach me. Thank you

Drawing function image with absolute value （1）y=|x-5|+|x+3| (2) Y = 2x-3 x ∈ Z | x | contained in Z （3）y=x²-2|x|-1 （4）y=｛x²+2x（x≥0） -x²-2x（x＜0） The best way to draw these four functions is to teach me. Thank you

1. Draw this section
When x ≤ - 3, y = - 2x + 2
When - 3 < x ≤ 5, y = - 2
When x > 5, y = 2x-2
Then draw the image according to the interval, which should be the shape of the pot
2. I think if x ∈ Z, | x | is absolutely contained in Z, which is a bit repetitive
The image is the point on the line y = 2x-3 whose abscissa is an integer
Three
Y = x & sup2; - 2 | x | - 1 can be regarded as y = | x | & sup2; - 2 | x | - 1
That is to say, the image of y = x & sup2; - 2x-1 on the right side of the Y axis is folded to the left with y as the symmetry axis
(you can also segment, when x > 0, y = x & sup2; - 2x-1
When x ≤ 0, y = x & sup2; + 2x-1)
4. What you give is already divided into sections. You can draw according to the interval
Talk in sections and draw a graph according to the X range
It is known that the center of the ellipse is at the origin, the focus is on the coordinate axis, the length of its major axis is three times that of its minor axis, and the ellipse passes through point a (3,1), and the elliptic equation is solved
It is known that the center of the ellipse is at the origin, the focus is on the coordinate axis, the length of its major axis is three times that of its minor axis, and the ellipse passes through point a (3,1), and the elliptic equation is solved
Let the focus of the ellipse be on the x-axis, and the equation is X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1, a (3,1) is on the ellipse, so there is an equation:
9/a²+1/b²=1.(1)
2A = 3 (2b), i.e. a = 3B. (2)
Substituting (2) into (1), we get 9 / (9b & # 178;) + 1 / B & # 178; = 1, so B & # 178; = 2, a & # 178; = 9b & # 178; = 18, so the elliptic equation is as follows:
x²/18+y²/2=1.
If the focus is on the y-axis, then let the elliptic equation be X & # 178 / B & # 178; + Y & # 178 / A & # 178; = 1, and then there is:
9/b²+1/a²=1.(3)
a=3b.(4)
Substituting (4) into (3), we get 9 / B & # 178; + 1 / 9b & # 178; = 1, so B & # 178; = 9 + 1 / 9 = 82 / 9, a & # 178; = 9b & # 178; = 82, so the elliptic equation is:
9x²/82+y²/82=1.
Functions with absolute values
1) Y = | X-5 | + | x + 3 | the analytical formula after the absolute value is removed
Y = | X-5 | + | x + 3 |: when x > = 5, y = 2x-2 when - 3
Urgent: the major axis is on the Y axis, the distance between the directrix is 36, and the distance between a point on the ellipse and the two focal points is 9 and 15 respectively?
If the distance from a point on the ellipse to two focal points is 9 and 15 respectively, then 2A = 9 + 15, a = 12. The distance between collimators = 2A & # 178 / C = 36, then C = 8, then B & # 178; = A & # 178; - C & # 178; = 80. The ellipse is Y & # 178 / 144 + X & # 178 / 80 = 1
Focus on the y-axis, let the elliptic equation be y / A ^ 2 + X / b ^ 2 = 1, because the distance between a point on the circle and two focus points is 24, so 2A = 24, and the distance between directrix is 36, so a ^ 2 / C = 18, a = 12, C = 8, and a ^ 2 = B ^ 2 + C ^ 2, then B ^ 2 = 80, so the equation is Y / 144 + X / 80 = 1
The alignment distance 2A ^ 2 / C = 36, 2A = 15 + 9, a = 12. C = 8. X ^ 2 / 80 + y ^ 2 / 144 = 0
The minimum positive period of the function y = 1-tan & # 178; 2x / 1 + Tan & # 178; 2x is
The deformation is as follows
y=1/2*2tan2x/(1-tan²2x)
=1/2*tan4x
The period of TaNx is π
So t = π / 4
The sum of the distances from a point P (3,2) on the ellipse to the two focal points is 8. Find the standard equation
2A = 8A = 4x & # 178 / 16 + Y & # 178 / B & # 178; = 19 / 16 + 4 / B & # 178; = 1b & # 178; = 64 / 7Y & # 178 / 16 + X & # 178 / B & # 178; = 14 / 16 + 9 / B & # 178; = 1b & # 178; = 12 so x & # 178 / 16 + 7Y & # 178; = 64 = 1x & # 178 / 12 + Y & # 178; = 1
1. The sum of the distances to the two foci is 8, so a = 4
2. Let x ^ 2 / 16 + y ^ / b ^ 2 = 1,
3. Substitution of P (3,2)
4. Get b
5. Get the equation
Finding the minimum positive period of function f (x) = (2tanx / 2) / (1-tan ^ 2x / 2)
f(x)=(2tanx/2)/(1-tan^2x/2)
=tanx
x/2≠kπ+π/2 ===>x≠2kπ+π ,x≠kπ+π/2
If we remove the undefined points, we can find that t = 2 π
T = π is wrong!
f(x)=tan(x/2+x/2)=tanx
PI
=tanx
The period is pi
When you change the tangent and cotangent to sine and cosine, apply the power reduction formula [SiNx] ＾ 2 = (1-cos2x) / 2
[cosx] ＾ 2 = (1 + cos2x) / 2 and 2cosxsinx = sin2x
Finally, TaNx is simplified
Then the period is Wu
I'll do it myself, and I won't do it for you again
The two focal coordinates are (- 4,0) and (- 4,0) respectively. The sum of the distances from one point P to two points on the ellipse is equal to 10
The two focal coordinates are (- 4,0) and (- 4,0) respectively, and C = 4
The sum of the distances from one point P to two points on the ellipse is equal to 10, so 2A = 10, a = 5
c^2=a^2-b^2
b^2=25-16=9
The equation is: x ^ 2 / 25 + y ^ 2 / 9 = 1
c=4,
2A = 10, so a = 5,
So B ^ 2 = a ^ 2-C ^ 2 = 25-16 = 9
So the elliptic standard equation is x ^ 2 / 25 + y ^ 2 / 9 = 1 (because the focus coordinate is on the X axis, so the elliptic equation is like this, it can't be y ^ 2 / 25 + x ^ 2 / 9 = 1)
Draw the absolute value of the function y = x & # / X
y=x²/|x|=|x|²/|x|=|x|    (x≠0)
First make the image of y = x, and then turn the image below the X axis symmetrically to the image above the X axis
According to X ≠ 0, remove the origin
Picture with upstairs, make good!
Write out the standard equation of the ellipse: the two focal coordinates are respectively [- 4,0]. (4.0), and the sum of the distances from a point P on the ellipse to the two focal points is equal to 0. [2] the two focal coordinates are respectively [0. - 2] (0.2) and pass through [- 3 / 2,5 / 2]