(2) given y = f (x) = x & # 178; - 2x + 3, when x ∈ [T, t + 1], find the maximum and minimum function values of the function

(2) given y = f (x) = x & # 178; - 2x + 3, when x ∈ [T, t + 1], find the maximum and minimum function values of the function

When t ≥ 1, f (x) max = f (T + 1) = T & # 178; + 2, f (x) min = f (T) = T & # 178; - 2T + 3; 2, when t + 1 ≤ 1, i.e. t ≤ 0, f (x) max = f (T) = T & # 178; - 2T + 3; 3, when 0

Find the maximum value of function y = - X & # 178; + 2x (t ≤ x ≤ T + 1)

① Find the axis of symmetry
x=-2/((-1)*2)=1
② Classification discussion:
1) When t + 1

Given the function y = x & # 178; + 2x-3, when - 4

Function y = x & # 178; + 2x-3 = (x + 1) &# 178; - 4
The opening of the image is upward, and the axis of symmetry x = - 1
When - 4

The function y = - 2x & # 178; + X has () value, what is the maximum value

The maximum value is one eighth

If the solution of the system of equations 4x-y equals 1 and Y equals 2x + 3 is x = 2 and y = 7, then the intersection coordinates of the images of the first-order functions y = 4x-1 and y = 2x + 3 are

Yes (2,7)
The root of binary linear equations is the coordinate of the intersection of two straight lines

2. As shown in Figure 1, it is known that the hyperbola Y1 = KX (k > 0) and the straight line y2 = k'x intersect at two points a and B, and the point a is in the first quadrant
2. As shown in Figure 1, it is known that the hyperbola Y1 = KX (k > 0) and the straight line y2 = k'x intersect at two points a and B, and point a is in the first quadrant
(1) If the coordinates of point a are (4,2), then the coordinates of point B are (- 4, - 2); when x satisfies: X ＜ - 4 or 0 ＜ x ＜ 4, Y1 ＞ Y2;
(2) Make another straight line L through the origin o, intersect the hyperbola y = KX (k > 0) at two points P and Q, and point P is in the first quadrant, as shown in Figure 2
① The quadrilateral apbq must be a parallelogram;
② If the coordinate of point a is (3,1) and the abscissa of point P is 1, calculate the area of quadrilateral apbq;
③ If the abscissa of points a and P are m and N respectively, can the quadrilateral apbq be a rectangle? If possible, the conditions for finding m and N; if not, please explain the reason

(1) Because both positive and negative scaling functions are centrosymmetric with respect to the origin, the coordinates of point B are B (- 4, - 2);
From the two functions passing through point a (4,2), we can see that the analytic formula of hyperbola is Y1 =, and that of straight line is y2 = X,
In each quadrant, the hyperbola y decreases with the increase of X, and the straight line y increases with the increase of X,
So when x ＜ - 4 or 0 ＜ x ＜ 4, Y1 ＞ Y2
(2) It is proved that both positive and inverse proportional functions are centrosymmetric with respect to the origin,
The results show that OA = ob, Op = OQ, and apbq must be a parallelogram according to the fact that the parallelogram bisected by the diagonal is a parallelogram
② The coordinates of point a are (3,1)
The hyperbola is y = 3 / X
So the coordinates of point P are (1,3),
The right angle trapezoid can be obtained by passing a as the vertical line of X axis, and then passing P as the vertical line of the vertical line,
By subtracting the area of the right triangle from the area of the right trapezoid, the area of the triangle POA is 4, and then the quadrilateral apbq is 16
③ When Mn = k, OA = OP, the quadrilateral with equal diagonals and bisecting each other is a rectangle, so the quadrilateral apbq is a rectangle
We are also working on this problem

As shown in Figure 1, it is known that the hyperbola Y1 = KX (k > 0) and the straight line y2 = k'x intersect at two points a and B, and point a is in the first quadrant
(1) If the coordinates of point a are (4,2), then the coordinates of point B are (4,2)______ When x satisfies:______ (2) make another straight line L through the origin o, intersect the hyperbola y = KX (k > 0) at two points P and Q, and point P is in the first quadrant, as shown in Figure 2______ (2) if the coordinate of point a is (3,1) and the abscissa of point P is 1, the area of the quadrilateral apbq can be calculated; (3) if the abscissa of points a and P are m and N respectively, can the quadrilateral apbq be a rectangle? If possible, find the conditions that m and n should satisfy; if not, explain the reason

(1) Because the positive proportion function and the inverse proportion function are centrosymmetric with respect to the origin, the coordinates of point B are B (- 4, - 2). From the two functions passing through point a (4,2), we can see that the analytic formula of hyperbola is Y1 = 8x, and that of straight line is y2 = 12x. In each quadrant, y of hyperbola decreases with the increase of X, and y of straight line increases with the increase of X, so when x ＜ - 4 or 0 ＜ x ＜ 4, Y1 > Y2 (2) ① both ∵ positive scale function and inverse scale function are centrosymmetric with respect to the origin, ∵ OA = ob, Op = OQ. According to the fact that the quadrilateral bisected by the diagonal is a parallelogram, apbq must be a parallelogram. ② the coordinates of ∵ a point are (3,1) ∵ the hyperbola is y = 3x, so the coordinates of P point are (1,3). If the perpendicular CD passing through a as X axis intersects X axis at C, the right angled trapezoid opdc, passing P can be obtained Let PD ⊥ DC and d be the perpendicular foot. The area of the triangle POA is 4 by subtracting the area of the right triangle from the area of the right trapezoid, and then the quadrilateral apbq is 16 by 4 × 4. ③ ∵ when Mn = k, then a (m, n), P (n, m), ∵ OA = OP, the quadrilateral with equal diagonals and bisecting each other is rectangular, and the quadrilateral apbq is rectangular

The left and right focus of the hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 are F1F2, the point P is on the right branch of the hyperbola, and Pf1 = 7pf2

It is known that the hyperbolic equation is: X & sup2; / A & sup2; - Y & sup2; / B & sup2; = 1  let the coordinates of P point be: (ASEC θ, btan θ) ∵ P point is on the right branch, so: - π / 2 < θ < π / 2 ∵ pf1-pf2 = 2A = 7pf2-pf2 = 6pf2 ∵ a = 3PF2 ∵ P: (ASEC θ, btan θ), F2 (C, 0) ∵ PF2 | & sup2; = (ASEC θ -...)

The left and right focus of the hyperbola x ^ 2 / 4-y ^ 2 / b ^ 2 = 1 is F1F2, and the point P is on the hyperbola, so that | Pf1 |, F1F2 |, | PF2 | is an arithmetic sequence, and | PF2 ||

So Pf1 + PF2 = 2f1f2 = 4C (1) and P is on hyperbola, so | pf1-pf2 | 2A = 4 (2) (1) ^ 2 + (2) ^ 2: Pf1 ^ 2 + PF2 ^ 2 = 2 (a ^ 2 + 4C ^ 2) O is the side of △ pf1f2, and the midpoint of F1F2 is determined by the conclusion: pf ^ 2 + PF2 ^ 2 = 2 (OP ^ 2 + of 1 ^ 2) SO 2 (a ^ 2

Hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, the left and right focus is F1F2, and there is a point P on the right branch, which satisfies
The left and right focus of hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 is F1 and F2, and there is a point P on the right branch, which satisfies: | op | = √ a ^ 2 + B ^ 2, if ∠ pf1f2 = ∠ pf2f1, what is the eccentricity of hyperbola?

Because | op | = √ a ^ 2 + B ^ 2 = C, so ∠ f1pf2 is right angle, and ∠ pf1f2 = ∠ pf2f1, so triangle pf1f2 is isosceles right triangle, so a = B, so a ^ 2 = B ^ 2 = C ^ 2-A ^ 2, so e = C / A = stem 2