The problem of the maximum value of quadratic function in the first grade of Higher Education 1. Find the maximum value of the function y = x Λ 2 + 2aX + 1, - 2 ＜ = x ＜ = 1. 2. Find the maximum value of the function y = x Λ 2 + 2x-1, m ＜ = x ＜ = m + 1. 3. Find the maximum value of the function y = - x Λ 2 + 2aX + 1-A, 0 ＜ = x ＜ = 1 is 2, and find a (x Λ 2 represents the square of x) (the answer should be detailed)

The problem of the maximum value of quadratic function in the first grade of Higher Education 1. Find the maximum value of the function y = x Λ 2 + 2aX + 1, - 2 ＜ = x ＜ = 1. 2. Find the maximum value of the function y = x Λ 2 + 2x-1, m ＜ = x ＜ = m + 1. 3. Find the maximum value of the function y = - x Λ 2 + 2aX + 1-A, 0 ＜ = x ＜ = 1 is 2, and find a (x Λ 2 represents the square of x) (the answer should be detailed)

one
y=x^2+2ax+1=(X+A)^2+1-A^2
Because - 2 ＜ = x ＜ = 1
Substituting 5-4a and 2 + 2a,
When 5-4a = 2 + 2a,
A=1/2
In this case, the two formulas are equal, and the value is y = x ^ 2 + 1,
Substituting - 2 ＜ = x ＜ = 1,
It is known that the maximum value of the function is [0,4]
So when a is less than 1 / 2, the maximum value of the function is [2 + 2a, 5-4a]
So when it is greater than 1 / 2, the maximum value of the function is [5-4a, 2 + 2A]
2.Y=X^2+2X+1
=(X+1)^2
By substituting m ＜ = x ＜ = m + 1, we can see that,
（M+1)^2

The maximum value of quadratic function
Given that the square of y = 4A (x-a) (a is greater than 0), and when x is greater than or equal to a, the minimum value of S = (x-3) ^ 2 + y ^ 2 is 4, find the parameter a

This topic seems very difficult, but its expression is too complex. Let's simplify it. Substitute y ^ 2 = 4A (x-a) into S (x) = (x-3) ^ 2 + y ^ 2 to get s (x) = (x-3) ^ 2 + 4a (x-a) = x ^ 2 + (4a-6) x + (9-4a ^ 2). It's not difficult to see that this is a quadratic function. Its axis of symmetry is x = 3-2a (and its open

It is known that y = y1-y2, Y1 is proportional to the square of X, and the scale coefficient is K1
It is known that y = y1-y2, Y1 is proportional to the square of X and the scaling coefficient is K1
Y2 is proportional to x + 3, and the proportional coefficient is K2
When x = 0, y = 2
When x = 3, y = 0
Find the functional relation between Y and X. is it a linear function?

y=y1-y2
y1=k1*x＾2
y2=k2*(x+3)
So y = K1 * x ＾ 2-k2 * (x + 3)
Substituting two known points, K1 = - 4 / 9, K2 = - 2 / 3
So the functional relation between Y and X is y = - 4 / 9x ＾ 2 + 2 / 3x + 2
Obviously it's not a linear function

Distribution of roots of quadratic function
On the equation x ^ 2 + ax + 2 = 0 of X, at least one real root is less than - 1, the range of a is obtained
It's better to do it in a variety of ways

The analysis of "at least one real root less than - 1" includes two cases: ① two real roots are less than - 1; ② one real root is less than - 1, and the other real root is not less than - 1
Solution 1 Let f (x) = x ^ 2 + ax + 2, the two real roots of the equation in the question are less than - 1, which is equivalent to
a^2-8>=0
-a/20
that is
a> = 2 times root 2 or a = 2
1-a+2>0
The solution
Double root 2

The two diagonals of diamond ABCD are 8 and 6 long respectively, point P is the moving point on diagonal AC, m and N are the midpoint of AB and BC respectively, then the minimum value of PM + PN is

5

Higher first quadratic function
It is known that the minimum value of F (x) = x & # 178; - ax + half a (a > 0) in the interval "0,1" is g (a) and the maximum value of G (a) is obtained
I'll see you tomorrow

f(x)=x²-ax+a/2=(x-a/2)²-a²/4+a/2
The function image is a parabola with the opening upward and the axis of symmetry x = A / 2
When 0

In rectangular ABCD, the diagonal lines AC and BD intersect at point O, ad = 6, CD = 8, P is the moving point on AB, PM is perpendicular to AC, PN is perpendicular to DB, and the value of PM + PN is obtained

From the meaning of the title
AO=BO=1/2BD
Because BD = under the root sign (AB & sup2; + BC & sup2;) = 10
So Ao = Bo = 5
Because the side length of △ ABO is 5,5,8
The height on the edge AB through o
On the Pythagorean theorem
The height is 3
So s △ ABO = 3 × 4 × 1 / 2 = 6
Connect op
Because s △ ABO = s △ AOP + s △ BOP=
1/2（AO×PM+BO×PN）=5/2（PM+PN）=6
So PM + PN = 2.4

If the image of quadratic function y = f (x) passes through the origin, and 1 is less than or equal to f (- 2) less than or equal to 2, 3 less than or equal to f (1) less than or equal to 4
25 / 3 less than or equal to f (2) less than or equal to 34 / 3

Because y = f (x) is a quadratic function, its equation can be set as y = f (x) = ax ^ 2 + BX + C
The image passes through the origin, i.e. the point (0,0)
So f (0) = ax0 ^ 2 + BX0 + C = 0, that is, C = 0
The quadratic function equation is reduced to y = f (x) = ax ^ 2 + BX
Another one

As shown in the figure, the diagonal of rectangle ABCD intersects at point O, P is any point of AD, PM ⊥ AC, PN ⊥ BD, if AB = 5, BC = 12, then PM + PN =?

1. Extend CD line, make parallel line of BD through a point, two lines intersect at e point, then bdea is parallelogram
2. When the NP line is extended and intersected with AE line at f point, NF is perpendicular to Ae
3. Because CD = De, ad is perpendicular to CE, so pf = PM. So PM + PN = NF
4. NF is the height of the right triangle abd, so NF = AB * BC / BD,
Calculation: ab = 5, BC = 12, so BD = 13
NF=5*12/13,
Results: NF = 60 / 13

In Cartesian coordinates, the parabola y = (- 4 / 9) x2 + (2 / 9) MX + (5 / 9) m + 4 / 3 intersects the x-axis with two points AB, point a is on the negative half axis, point B is on the positive half axis, OB = 2oa (with absolute value), and point C is the vertex of the parabola. Find the value of M

M is 4 or (- 3 / 2)