The known function f (x) = 2x-1 g (x) = {X & # 178;, X ≥ 0} (piecewise function) - 1, x < 0 Why do we need to find the norm of X again to find the analytic expression of G [f (x)] g（x）={x²，x≥0 } -1，x＜0

The known function f (x) = 2x-1 g (x) = {X & # 178;, X ≥ 0} (piecewise function) - 1, x < 0 Why do we need to find the norm of X again to find the analytic expression of G [f (x)] g（x）={x²，x≥0 } -1，x＜0

Analysis: first of all, replace all x in G (x) with F (x), write out the expression of G [f (x)], and then solve. G [f (x)] = [f (x)] &# 178;, f (x) ≥ 0 (because now f (x) replaces the independent variable of X in function g (x), which is simply the position) - 1

Is the function graph of y = a (X-H) ² + k the same as the function graph of which formula and with the same opening?
How does y = a (X-H) &# 178; + K transform from y = ax & # 178

Translation: left plus right minus on X, up plus down minus on y
Y = ax & # 178; shift h unit to the right
That is y = a (X-H) ²
And then up the k-unit
That is y = a (X-H) ² + K
If you agree with my answer, please click "adopt as a satisfactory answer" in the lower left corner. I wish you progress in your study!

Given that point P is the point on the left branch of hyperbola C: x2a2 − y2b2 = 1 (a > 0, b > 0), F1 and F2 are the left and right focuses of hyperbola, and Pf1 ⊥ PF2, PF2 and two asymptotes intersect at two points m and n (as shown in the figure), and point n exactly bisects segment PF2, then the eccentricity of hyperbola is ()
A. 5B. 2C. 3D. 2

In the triangle f1f2p, the point n exactly bisects the line segment PF2, the point O exactly bisects the line segment F1F2, | on ‖ Pf1, and the slope of on is Ba, | Tan ∠ pf1f2 = ba. In the triangle f1f2p, let PF2 = BT. Pf1 = at. According to the definition of hyperbola, | PF2 | - | Pf1 | = 2A, | BT at = 2A

If the line y = x + m and hyperbola y = m / X intersect at point a in the first quadrant and the X axis intersects at point B, AC is perpendicular to the X axis, and the vertical axis is point C, and the area of AOC is 3, the solution of M is obtained

M=3*2=6

A ` B is two points on the hyperbola, the abscissa of a + + B is 1 + + 2, the extension of line AB intersects the X axis at point C. if the area of △ AOC is 6, the value of K is calculated?

Let hyperbola y = K / x, then a (1, K) B (2, K / 2) has ab: y = (- K / 2) x + (3 / 2) K
C(3,0)
And because the area of △ AOC is 6, so a (1,4)
The value of K is 4

As shown in the figure, a and B are the points on the hyperbola y = KX (k > 0). The abscissa of a and B are a and 2A respectively. The extension line of line AB intersects the X axis at point C. if s △ AOC = 6, then the value of K is ()
A. 1b. 2C. 4D. Not sure

Then ad ∥ be, ad = 2be = Ka, ∥ B, e are the midpoint of AC and DC respectively. ∥ ADC ∥ BEC, ∥ be: ad = 1:2, ∥ EC: CD = 1:2, ∥ EC = de = a, ∥ OC = 3A, and ∥ a (a, KA), B (2a, k2a), ∥ s ∥ AOC = 12ad × C

A. B is the point on the hyperbola y = K / X (k.0). The abscissa of a and B are a and 2A respectively. The extension of line AB intersects the X axis at point C. if s △ AOC = 6, the value of K is calculated
Make X-axis through a, B and E, f respectively. BF is half of AE obviously. Why?

Then ad ∥ be, ad = 2be = Ka, ∥ B, e are the midpoint of AC and DC respectively. ∥ ADC ∥ BEC, ∥ be: ad = 1:2, ∥ EC: CD = 1:2, ∥ EC = de = a, ∥ OC = 3A, and ∥ a (a, KA), B (2a, k2a), ∥ s △ AOC = 12, ad × co = 12 × 3a × Ka = 3k2 = 6, the solution is k = 4

As shown in the figure, the line y = - x + 5 and the hyperbola y = 4 / X intersect at two points a and B, and point C is a point between a and B on the hyperbola, then the maximum area of △ ABC is

Analysis: when the area of △ ABC is the largest, the distance from point C to AB is the largest, so point C must be the tangent point parallel to the straight line AB and tangent to the hyperbola. Tangent means that there is only one intersection point between the straight line and the hyperbola. Let the analytical formula of the straight line be: y = - x + B, substituting y = 4 / x, we can get: - x + B = 4 / XX & # 178; - BX + 4 = 0 △ = B & # 178; -

Let ABC be an equilateral triangle, then the eccentricity of the hyperbola with a and B as the focus and passing through the midpoint of BC is?

Let the length of the regular triangle be 2 and the focus of the hyperbola be on the x-axis,
2C = 2 defined by hyperbola, that is, C = 1;
If you go through B and C, the midpoint is set to P, because it is an equilateral triangle and AP is three lines in one,
So the triangle APB is a right triangle, and we can find that AP = root 3, BP = 1,
The difference between the moving point (P) and the fixed point (a, b) defined by hyperbola is a constant (2a),
So PA Pb = 2A = radical 3-1;
So the eccentricity e = C / a = 2 / (radical 3-1), the denominator is rational

The straight line y = x + 5-6m intersects the hyperbola y = M-X at the point a of the first quadrant, intersects the x-axis at the point C, AB is perpendicular to the x-axis at the point B, if the s triangle AOB = 3, the
The straight line y = x + 5-6m intersects the hyperbola y = M-X at the point a of the first quadrant, intersects the x-axis at the point C, AB is perpendicular to the x-axis at the point B, if the s triangle AOB = 3,
Then s triangle AOC =?

Let a (x1, Y1), then s triangle AOB = 1 / 2 * X1 * Y1 = 3, so X1 * Y1 = 6, and a is on hyperbola, so X1 * Y1 = M = 6. The linear and curvilinear equations form a system of equations, and the coordinates of point a (1,6) are solved,
S triangle AOC = s triangle abc-s triangle AOB = 1 / 2 * 6 * 6-3 = 15