What is the area s of the plane figure enclosed by the second power of the curve X = y and y = x

What is the area s of the plane figure enclosed by the second power of the curve X = y and y = x

X=y
y=x²
The intersection point is (0,0) (1,1)
s=∫（0,1）(x-x²)dx
=1/2x²-1/3x³|（0,1）
=1/2-1/3
=1/6

Through P (1,0), make the tangent line under the parabola y = root sign (X-2). The tangent line forms a plane figure with the above parabola and X axis. Try to find the area of the plane figure

Y '= - 1 / [2 √ (X-2)], set the tangent point coordinates P (x0, Y0),
(y0-0)/(x0-1)=1/[2√（x0-2)],
y0=√（x0-2),
[√（x0-2)]/(x0-1)=1/[2√（x0-2)],
2x0-4=x0-1,
x0=3,y0=1,
The tangent point coordinates P (3,1),
Tangent equation: (y-0) / (x-1) = 1 / 2,
y=x/2-1/2,
The graph area is surrounded by the curve y = x / 2-1 / 2, y = √ (X-2) and X axis,
For a straight line X coordinate value is [1,3], for a parabola [2,3]
S=∫[1,3][x/2-∫[2,3]√(x-2)]dx
=[x^2/4][1,3]-(2/3)(x-2)^(3/2)][2,3]
=（9-1）/4-（2/3）*（1-0）
=4/3.

When the image of y = cosx root 3 times SiNx is translated along the direction of vector a = (- m, m) m > 0, the minimum value of M of the image obtained is symmetric about the y-axis

y=cosx-√3sinx
=2cos(x+π/3)
The function after translation along vector a = (- m, m) is
y=2cos(x+m+π/3)+m
∵ this function is symmetric about the y-axis
∴2cos(-x+m+π/3)+m=2cos(x+m+π/3)+m
ν - x + m + π / 3 = x + m + π / 3 or - x + m + π / 3 + X + m + π / 3 = 2K π, where k is an integer
ν x = 0 or M + π / 3 = k π, where k is an integer
∵m>0
ν take k = 1, and M = 2 π / 3
The minimum value of M is 2 π / 3
The second way of thinking is
If the function y = 2cos (x + m + π / 3) + m is symmetric about the y-axis,
Then M + π / 3 = k π, K is an integer. Combining with M > 0, the minimum value of M is 2 π / 3

If the image of the function f (x) = root 3cosx SiNx is translated by vector (- A, 0) (a > 0), the corresponding function of the image is even function, then the minimum value of a is

F (x) = (√ 3) cosx - SiNx = 2 [(√ 3 / 2) cosx - (1 / 2) SiNx] = 2 [sin (π / 3) cosx - cos (π / 3) SiNx] = 2Sin (π / 3 - x) = - 2Sin (x - π / 3) the image is translated by vector (- A, 0), which is equivalent to a unit of left translation

The quadratic function y = 2x ^ 2-8x + 3 is known. The parabola analytic formula of x-axis symmetry with the image of this function? If you will, please analyze it The quadratic function y = 2x ^ 2-8x + 3 is known. The parabola analytic formula of x-axis symmetry with the image of this function? Please analyze it if you will What does "about X-axis symmetry" mean?

By simplifying y = 2 (x ^ 2-4x + 4) - 5, y = 2 (X-2) ^ 2-5, we can see that the vertex coordinates are (2, - 5), and the opening is upward. Let y = 0, x = 0 to get the value, which is the intersection point of the image with X axis and Y axis!

If the image of the quadratic function y = x ^ (2) + BX + C passes through the origin and is symmetric about the y-axis, then the analytic formula of the quadratic function is?

Through the origin, we know that C = 0, the opening is upward, and the drawing about Y-axis symmetry can only be y = x ^ 2

Corresponding to the image of the quadratic function y = 1 / 2x squared + 2 with respect to the x-axis symmetry, and its function analytic formula (step-by-step)

Let (x, y) be any point in the graph of the function to be solved, and its point (x, - y) which is symmetric about the X axis on the image of the quadratic function y = 1 / 2x squared + 2 is substituted into - y = x ^ 2 / 2 + 2, that is, y = - x ^ 2 / 2-2

What are the functional relations of quadratic function y = - 2 (x-1) ^ 2 + 1 about origin, x-axis and y-axis symmetry?

Origin: - x instead of X, - y instead of Y, y = 2 (x + 1) ^ 2-1
X axis: - y instead of Y Y Y = 2 (x-1) ^ 2-1
Y axis: - x instead of x y = - 2 (x + 1) ^ 2 + 1

What are the functional relations of quadratic function y = x ^ 2-4x + 1 about origin, X axis and Y axis symmetry?

The function relation of origin symmetry is: - y = (- x) ^ 2-4 (- x) + 1, that is, y = - x ^ 2-4x-1
The functional relation of x-axis symmetry is: - y = x ^ 2-4x + 1, that is, y = - x ^ 2 + 4x-1
The function relation of y-axis symmetry is: y = (- x) ^ 2-4 (- x) + 1, that is, y = x ^ 2 + 4x + 1

Known quadratic function y = x2-2mx + M2 + m-2 (1) When m is the value, the image of the quadratic function passes through the origin (2) When m is the value, the image of quadratic function is symmetric about y axis

(1) ∵ the image of the quadratic function y = x2-2mx + M2 + m-2 crosses the origin,
 substituting (0, 0) into, we get: M2 + m-2 = 0,
M = 1 or - 2,
Therefore, when m is 1 or - 2, the image of quadratic function passes through the origin;
(2) ∵ the symmetry axis of quadratic function is y-axis,
∴-2m=0，
M = 0
Therefore, when m is 0, the image of quadratic function is symmetric about y-axis