Given that the vertex coordinates of the parabola are (- 2,1) and pass through points (1, - 2), find the functional expression of the parabola

Given that the vertex coordinates of the parabola are (- 2,1) and pass through points (1, - 2), find the functional expression of the parabola

Let y = a (x + 2) ²+ one
Bring in (1, - 2)
-2=9a+1
a=-1/3
y=-1/3（x+2） ²+ one
=-1/3x ²- 4/3x-1/3

How to prove that a function is bounded (f) x = x ² sinx

It can be proved that this function is unbounded
When x = 2K π + π / 2, f (x) = (2k π + π / 2) ²
When k - > + ∞, f (x) - > + ∞
Therefore, the function is unbounded

Find the function y = x2+9+ X2 − 10x + 29 min

∵y=
x2+9+
x2-10x+29，
∴y=
(x-0)2+(3-0)2+
(x-5)2+(0+2)2，
It can be regarded as the sum of the distances from the moving point P (x, 0) on the X axis to the two fixed points a (0, 3) and B (5, - 2),
Known from "the shortest line segment between two points",
When points a, P and B are collinear,
That is, when x = 3, Ymin = |ab| = 5
2．
So function y=
x2+9+
The minimum value of x2-10x + 29 is 5
2．

Kneel down and seek the master of mathematics ~ ~ ~ find the value range ① y = 3-2sin2x ② y = |sinx| + SiNx ③ y = 1-1 / 2 SiNx under the root sign. The steps to be detailed~ And, and Known function y = 1 + 2sinx, when x=_____ When, y takes the maximum value _; When x=____ When, y goes to the minimum____ Known function y = 1-2sinx, when x=_____ When, y takes the maximum value _; When x=____ When, y goes to the minimum____ Thank you ~ ~!

My God, so many questions... You don't want to do your homework
1. Because the value range of sin2x is [- 1,1]
Therefore, the value range of y = 3-2sin2x is [1,5]
2. Segment discussion
When SiNx

When x belongs to [- π / 2, π / 2], the maximum value of function f (x) = SiNx + radical 3cosx?

f(x)=sinx+√3cosx
=2﹙sinxcosπ/3+cosxsinπ/3﹚
=2sin﹙x＋π/3﹚
∵ x belongs to [- π / 2, π / 2]
∴x＋π/3∈[-π/6,5π/6】
∴f(x)=sinx+√3cosx
Where x + π / 3 = - π / 6, i.e
When x = - π / 2, take the minimum value of 2 × ﹙﹣ ½ ) = - 1, where x + π / 3 = π / 2
When x = π / 6, the maximum value is 2 × 1=2

How to find the number of roots of cubic equation by derivative method? For example, x ^ 3-6x ^ 2 + 9x-10 = 0

Let f (x) = x ^ 3-6x ^ 2 + 9x-10, then the derivative of F (x) is 3x ^ 2-12x + 9 = 3 (x-3) (x-1)
When x = 3 or x = 1, the derivative is 0
When x > = 3 or X=