What is the maximum or minimum value of y = a (x-C) &#?

What is the maximum or minimum value of y = a (x-C) &#?

(x-c)²>=0
So Y > = 0

Find the maximum and minimum of y = x & # 178; + 1 on [- 2,1]

Minimum = y (0) = 1
Maximum value = (- 2) &# 178; + 1 = 4 + 1 = 5

If x and y satisfy X & # 178; + Y & # 178; = 1, then the minimum value of (Y-2) / (x-1) is; the maximum value of X / 3 + Y / 4 is

1. (Y-2) / (x-1) can be regarded as the slope of the line between any point (x, y) and point (1,2) on a circle, so it is easy to draw a graph. When the minimum value is taken, the tangent line passing through point (1,2) is tangent to the circle. Let y = K (x-1) + 2, the distance from the center of the circle to the straight line d = | - K + 2 | / √ (K & # 178; + 1) = 1, and the solution is k = 3 / 4, which is the minimum value of 3 / 4

If we know that the point of complex z = 4 + 2I (1 + I) 2 (I is an imaginary unit) in the complex plane is on the straight line x-2y + M = 0, then M = ()
A. -5B. -3C. 3D. 5

∵ the point corresponding to the complex z = 4 + 2I (1 + I) 2 = 4 + 2i2i = 2 + II = − I (2 + I) − I · I = 1-2i is (1, - 2). Substituting the line x-2y + M = 0, we can get 1-2 × (- 2) + M = 0, and the solution is m = - 5