Y = (log base 3 x) square - (2log base 3 x) + 2, where x ∈ [1 / 3,9], find the range

Y = (log base 3 x) square - (2log base 3 x) + 2, where x ∈ [1 / 3,9], find the range

y=（log₃x)²-2log₃x+2
=(log₃x-1)²+1
∵x∈[1/3,9]
∴log₃1/3≤log₃x≤log₃9
∴-1≤log₃x≤2
When log &; X = 1, Ymin = 1
When log &; = - 1, ymax = 5
The value range of the function is [1.5]

Log 3 ^ 2 with 2 as the base = 2log 2 ^ 3, why? How?

This is the formula, loga B ^ n = nloga B

Log takes 9 as the base and 27 as the true number=

Log base on 9 27 = log base on 9 (9 × 3)
=9 is the bottom 9 in log + 9 is the bottom 3 in log
=1+1/2
=1 and 1 / 2
=3/2
If you understand and solve your problem, please accept it as a satisfactory answer in time. Please understand,

25. As shown in the figure, it is known that the parabola y = - x2 + BX + C passes through point a (2,0), the symmetry axis is Y axis, and the vertex is p. (1) find the expression of the parabola
25. As shown in the figure, it is known that the parabola y = - x2 + BX + C passes through point a (2,0), the symmetry axis is y-axis, and the vertex is p
(1) Find the expression of the parabola, write out the coordinates of its vertex P, and draw its general image;
(2) Translate the parabola m units to the right and then M units to the down (M > 0). Note that the vertex of the new parabola is B and the intersection of the new parabola and the Y axis is C
(3) The coordinates of point B and point C are expressed by the algebraic expression of M. if ∠ OBC = 45 °, try to find the value of M
The key is the second question of (3)

Analysis: (1) the idea of solving the problem is to substitute the coordinates and axis of symmetry (x = 0) of point a into the parabola y = - x2 + BX + C to get the expression and vertex coordinates;
(2) According to the law of translation, we can get the coordinates of the vertex B of the new parabola and the intersection C of the new parabola and the y-axis;
② The small problem is to prove that two triangles are similar, and then use the ratio of the sides of similar triangles to be equal to get the value of M
(1) ∵ the parabola y = - x2 + BX + C passes through point a (2,0), and the symmetry axis is Y axis
｛0=-4+2b+c;-b/2×（-1)=0
∴b=0,c=4,
∴y=-x2+4,
When x = 0, y = 4,
The coordinates of P are (0,4),
So: the expression of the parabola is: y = - x2 + 4, and the coordinates of its vertex P are: (0,4)
(2) (1) the parabola first shifts m units to the right and then M units to the down (M > 0)
∴B（m,4-m）,
∵y=-（x-m）2+4-m,
When x = 0, y = - m2-m + 4,
∴C（0,-m2-m+4）,
So, using the algebraic expression of M, the coordinates of point B are: (m, 4-m), and the coordinates of point C are: (0, - m2-m + 4)
② B as BN ⊥ Y axis in N,
∵ as we know, the parabola first moves m units to the right, and then moves m units down,
∴PN=BN=m,∠BNP=90°
In this paper, we discuss the relationship between OPB and OBC,
And ∠ POB = ∠ POB,
The OCB is similar to the OBP
When point C is on the positive half axis of y-axis, i.e. - m2-m + 4 ＞ 0, Bo2 = OC &; op,
∵BO2=2m2-8m+16,OC=-m2-m+4,OP=4．
The solution is M1 = 0 (rounding off), M2 = 2 / 3
In addition, through point C, CD ⊥ ob is at point D, and through point B, be ⊥ OC is at point E,
Similarly, using △ OCD ∽ OBE
When point C is on the negative half axis of Y-axis and point - m2-m + 4 ＜ 0, BC2 = OC &; CP,
∵BC2=m2+m4,OC=m2+m-4,CP=m2+m．
The solution is M1 = 0 (rounding off), m2,3 = 1 ± √ 3 (rounding off the negative root)
∴m=1+√3
So the value of M is 2 / 3 or 1 + √ 3

There are four scores: 11 / 24 12 / 25 11 / 29 19 / 39, in which the difference between the largest and the smallest is equal to________ .

122/1131

If a = A2 + 5b2-4ab + 2B + 100, find the minimum value of A

The minimum value of a = a2-4ab + 4B2 + B2 + 2B + + 1 + 99 = (a-2b) 2 + (B + 1) 2 + 99, ∵ (a-2b) 2 ≥ 0, (B + 1) 2 ≥ 0, ∵ a ≥ 99, ∵ A is 99

How to find ((3 + X & # 178;) ^ 0.5) '?

（3+x²）^(1/2)
（（3+x²）^(1/2)）’
=1/2*(3+x²)^(1/2-1)*(3+x²)'
=1/[2√(3+x²)]*(0+2x)
=2x/[2√(3+x²)]
=x/√(3+x²)

Factorization: ab (a-b) + BC (B-C) + Ca (C-A)

The first volume of junior high school mathematics Chapter 4 equation of one variable, from problem to equation
If Party A works alone for 4 days, and Party B works alone for 6 days, if Party A works for one day, and then Party A and Party B cooperate to complete the work, if Party A works for X days, the number of days for Party B is______ From this equation can be listed___________ .

x-1
1/4*x+1/6*(x-1)=1

Why is a symmetric matrix positive definite if and only if all eigenvalues are greater than 0?

The orthogonality of real symmetric matrix is similar to that of diagonal matrix
That is to say, it is congruent with diagonal matrix
The elements on the main diagonal of diagonal matrix are the eigenvalues of A
So the eigenvalues of the symmetric matrix A are all greater than 0