Given that the symmetry axis of quadratic function y = x ^ 2-bx + 3 is x = - 2, find the value of B and vertex coordinates

Given that the symmetry axis of quadratic function y = x ^ 2-bx + 3 is x = - 2, find the value of B and vertex coordinates

The symmetry axis of y = x ^ 2-bx + 3 is x = - (- b) / 2 = B / 2 = - 2
b=-4
y=x^2+4x+3
x=-2,y=4-8+3=-1
So the vertex (- 2, - 1)

The vertex coordinates of the quadratic function y = x square + 1 are

y=x^2+1
The vertex coordinates are (0,1)

The vertex coordinates of the image of the quadratic function y = x + 2 are

（0,2）

The length and width of the rectangle are calculated by folding 22cm long wire into 30cm 2 rectangle

Let the length of the rectangle be xcm, (1 point) according to the meaning x (222-x) = 30, we can get x2-11x + 30 = 0, (6 points) solve the equation, we can get X1 = 5, X2 = 6, from X1 = 5 we can get 222-x = 6 (not consistent with the problem, omit). From x2 = 6 we can get 222-x = 5. (11 points) answer: the length of the rectangle is 6cm, the width is 5cm. (12 points)

Use a 30cm long wire to circle a rectangle with a length to width ratio of 3:2. What's its area?

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Length: 30 △ 2 × 3 / (3 + 2) = 9 (CM)
Width: 30 △ 2 × 2 / (3 + 2) = 6 (CM)
Area: 9 × 6 = 54 (square centimeter)

A 100cm long iron wire is divided into two parts, each part is bent into a square, and their area sum is the smallest______ cm2．

Let the wire be divided into two parts xcm and (100-x) cm, and the equation is: y = (x4) 2 + (100 − x4) 2 = 18 (x-50) 2 + 312.5. From the properties of the function, we know that because 18 > 0, the minimum value is 312.5cm2. So the answer is: 312.5

If the 12 cm long thin wire is cut into two sections and each is surrounded by an equilateral triangle, then the minimum value of the sum of the two equilateral triangles is ()
A. 323cm2B. 4cm2C. 32cm2D. 23cm2

Let the length of the two regular triangles be xcm, (12-x) cm respectively, then the sum of the areas of the two regular triangles is s = 34 (x3) 2 + 34 (12 − x3) 2 = 318 (x2-12x + 72) = 318 [(X-6) 2 + 36] ≥ 23, so D

A 120cm long wire is divided into two parts, each part is bent into a square. What is the sum of their areas? What's the area and minimum of them?

Let the iron wire be divided into two parts xcm and (120-x) cm, and the sum of the areas is y. The equation shows that y = (x4) 2 + (120 − x4) 2 = 18 (X-60) 2 + 450 ∥ the sum of their areas is 18 (X-60) 2 + 450, and the minimum value is 450cm2

Divide the 12 cm iron wire into two sections to form an equilateral triangle. What is the minimum value of the sum of their areas?

The area of an equilateral triangle with side length a = the quotient of root 3 divided by 4 times the square of A
It is (3 ^ 0.5 / 4) * a ^ 2, which is proportional to the square of the side length
Let two sides be a and B respectively,
Because a ^ 2 + B ^ 2 is greater than or equal to (a + b) ^ 2 / 2, if and only if a = B, the equal sign holds. Therefore, when the 12 cm iron wire is divided into two 6 cm segments, the area sum of the equilateral triangles is the smallest, and the minimum value is 18 * 3 ^ 0.5
Note: x ^ y represents the power of X to y, for example, 2 ^ 3 = 8,2 ^ 0.5 = 1.414

A 120cm long wire is divided into two parts, each part is bent into a square. What is the sum of their areas? What's the area and minimum of them?

Let the iron wire be divided into two parts xcm and (120-x) cm, and the sum of the areas is y. The equation shows that y = (x4) 2 + (120 − x4) 2 = 18 (X-60) 2 + 450 ∥ the sum of their areas is 18 (X-60) 2 + 450, and the minimum value is 450cm2