The minimum and maximum of y = x square-2x-3

The minimum and maximum of y = x square-2x-3

Y = x square-2x-3 = (x-1) ^ 2-4 when x = 1, the minimum value is - 4, because the coefficient of quadratic term > 0, there is no maximum value

If the definition field of function y = a x power (a greater than 1) is [negative one, 1] and the difference between the maximum and minimum is 1, what is a equal to

a>1
So it's an increasing function
So the biggest is a
The minimum is - 1 power of a = 1 / A
So A-1 / a = 1
a²-a-1=0
And a > 1
So a = (1 + √ 5) / 2

X square plus y square equals one, find the minimum value of root sign 3x plus y

Let √ 3x + y = a
y=a-√3x
So x & # 178; + Y & # 178;
=x²+a²-2√3ax+3x²=1
4x²-2√3ax+(a²-1)=0
X is a real number
So △ > = 0
12a²-16a²+16>=0
a²

In the interval [- 2,5], when x is equal to, what is the maximum value; when x is equal to, what is the minimum value
Please give me a detailed explanation

When 2 is less than or equal to X is less than or equal to 5,
When y = x ^ 2 -- 3x -- 4, x = 2, there is a minimum y = -- 6
When x = 5, there is a maximum y = 6
When -- 2 is less than or equal to X is less than or equal to 2,
When y = -- X ^ 2 -- 3x + 4, x = -- 3 / 2, there is a maximum y = 25 / 4
When x = 2, there is a minimum y = -- 6
When x = -- 3 / 2, the maximum value is 25 / 4,
When x = 2, take the minimum value -- 6

If △ ABC is equal to △ EFG, ab = EF, BC = FG, ∠ a = 68 ° and ∠ F - ∠ g = 56 °, then ∠ g = how much and ∠ B = how much

∵△ABC≌△EFG,AB=EF,BC=FG
∴∠E=∠A=68°
∠F+∠G=180°-68°=112°
And ∠ F - ∠ g = 56 °
The solution is ∠ g = 28 ° and ∠ f = 84 °
∠B=∠F=84°

It is known that | a | = 8 and E is the unit vector. When the angle between them is π 3, the projection of a in the direction of E is ()
A. 43B. 4C. 42D. 8+23

From the geometric meaning of the scalar product of two vectors, we can see that the projection of a in the direction of E is a · e = | a | e | cos π 3 = 8 × 1 × 12 = 4, so we choose B

As shown in the figure, it is known that in △ ABC, ∠ B = 90 °, ab = BC, BD = CE, M is the middle point on the edge of AC. it is proved that point m is on the vertical bisector of line segment De

Connecting BM;
It is easy to know that △ ABC is an isosceles right triangle
The two base angles of isosceles right triangle are equal and equal to 45 degrees
BM is the center line on the hypotenuse of an isosceles right triangle
∴AM=BM=CM
℅ ∠ ABM = ∠ a = 45 ° (equal sides and equal angles)
∵∠ABM=45°,∠C=45°,BM=CM,BD=CE
≌△ MDB ≌△ mec (two congruent triangles with equal angles on both sides)
Ψ MD = me (the corresponding sides of congruent triangles are equal)
A DEM is an isosceles triangle (a triangle with equal sides is called an isosceles triangle)

The distance from a point P to the left focus is 12 on the square of ellipse X / 100 + the square of Y / 36 = 1. What is the distance from it to the directrix of the ellipse?
What is the second definition?

a=10 b=6 c=8 e=4/5
So the distance from P to the right focus is 8
Because the ratio of the distance from P to the right focus to the distance from P to the right guide line is equal to the eccentricity e (second definition)
So the distance from P to the right guide line is 10
The second definition: the locus of a point whose ratio of the distance to a fixed point and the distance to a fixed line is a normal number is called a conic curve. The fixed point is called the focus, the fixed line is called the Quasilinear, and the normal number is called the eccentricity. If the eccentricity is greater than 1, the curve is hyperbolic; if the eccentricity is equal to 1, the curve is parabolic; if the eccentricity is less than 1, the curve is hyperbolic, The ellipse has two focal points (left focal point and right focal point) and two guide lines (left guide line and right guide line). The ratio of the distance from the point on the ellipse to the left (right) focal point and the distance to the left (right) guide line is equal to the eccentricity

As shown in the figure, in △ ABC, ab = AC, BC = BD = ed = EA, find the degree of ∠ a=______ ．

∵AE=ED，∴∠ADE=∠A，∴∠DEB=∠A+∠ADE=2∠A，∵BD=ED，∴∠ABD=∠DEB=2∠A，∴∠BDC=∠A+∠ABD=3∠A，∵BD=BC，∴∠C=∠BDC=3∠A，∵AB=AC，∴∠ABC=∠C=3∠A，∵∠ABC+∠C+∠A=180°，∴7∠A=180°，∴∠A=180...

The definition field of the function y = KX + 7 / KX ^ 2 + 2kx + 1 + k is r, and the value range of K is obtained

The definition field of the solution function y = KX + 7 / KX ^ 2 + 2kx + 1 + k is r, that is, KX ^ 2 + 2kx + 1 + K ≠ 0 is r constant for X. when k = 0, KX ^ 2 + 2kx + 1 + k = 1 ≠ 0 is R constant for X. when k ≠ 0, it is known that Δ < 0 is (2k) &# 178; - 4 * k (1 + k) < 0 is 4K & # 178; - 4K & # 178