Ask a question about mathematical quadratic functions!

1) Solution: Let y=kx+b

When x=, y=; When x=1, y=2 is substituted:

2=k+b is solved: k=-1, b=3

So, y=-x+3

2) Profit = (Unit Price - Unit Cost) * Sales Volume.

w=(w=(

Dissolve: w=-x 2+

When x=2, w=

1) Because it is a primary function, let y=kx+b, then, k+b=2 gives k=-1, b=3, so the analytical formula is y=-x+3

2) The functional relationship between w and x is w=(-x+3) (, when x=2, w = 10,000 yuan.

Solution: (1) Let the primary function be y=kx+b

Then there is: k+b=2 solution: k=-1, b=3

The function relationship is: y=-x+3

2) w=(when x=2, w=10,000 yuan.)

Solution: (1) y=2x +4x-6

2(x^2+2x)-6

2(x^2+2x+1-1)-6

2(x+1) 2-2-6=2(x+1) 2-8(2) because a=2 0, h=-1, k=-8

So the opening is upward, and the axis of symmetry is x=-1

The vertex coordinates are (-1, -8).

3) Because the image has an intersection point with the x-axis, y=0

then y=2x +4x-6=0

The solution yields x=-3 or x=1

Therefore, the coordinates of the intersection of the image and the two coordinates (-3,0)(1,0)(4) The image is shifted by one unit to the left by parabola y=2x, and then 8 units downwards (6) When x<=-1, y decreases with the increase of x.

7）.Y > 0 at x>1 or x<-3, y=0 at x=1 and -3, and -3(8) when x=-1, the minimum value of y is -8

2.The opening is up, the axis of symmetry x=-1, the vertex coordinates (-1, -8) 3Intersection with the x-axis (-3,0)(1,0), intersection with the y-axis (0,-6)4Translate 1 unit to the left and 8 units down;

6.When x <-1, y decreases as x increases.

7.When -13 or x<-1, y>0

8.When x=-1, the function y has a minimum value. The minimum value is -8.

2。The opening is up, the axis of symmetry x=-1, the vertex coordinates (-1, -8) 3The intersection points with the x-axis are (-3,0) and (1,0), and the intersection points with the y-axis (0,-6).

4。It is obtained by y=2x by 1 unit to the left and then 8 units down.

6。When x -1 y decreases as x increases.

7。Actually, y=2(x-1)(x+3), so y=0 when x=1 or x=-3, and y 0 when -3 x 1;When x -3 or x 1 y 0

8。When x = -1 there is a minimum value of -8,.

Okay, it's over, go ahead, these are very basic questions, so it seems that your function needs to be taken to the next level.

（1）y=2(x+2)^2-10

2) The opening is upward, the axis of symmetry x=-2, the vertex (-2, -10) (3) the intersection point with the x-axis: (-3, 0), (1, 0); Intersection with y-axis: (0,-6)(4)The function is translated by 2 units from y=2x 2 and 10 units downwards to obtain (6) When x <-2, y decreases as x increases.

7) When x<-3 or x>1, y>0;When x=-3 or x=1, y=1;When -3(8) when x = -2, y has a minimum value of -10

(1) y=2(x+1) 2-8, (2) the opening is upward, the axis of symmetry is the straight line x=-1, the vertex (-1, -8) (3) and the intersection point of the x-axis a(1,0)b(-3,0) and y-axis (0,-6)(4) are the parabola y=2x 2 by translating one unit to the left, and then translating 8 units downwards to obtain y=2x 2+4x-6(6) take x<-1, (7) when x<-3 or x>1 y>0, when x takes -3 or 1 y=0, when -3 thank you for adopting!

2.The opening is upward, the axis of symmetry x=-1, the vertex (-1, -8) 3with y-axis (0,-6), with x-axis (1,0),(3,0)4

From the parabola y=2x, first want to translate 1 unit to the left, and then 8 units to the downward 6When x<=-1.

y<0 at 1 or x>-3, y=0, -3 at x=1 and -3< x < 1 when y < 0, y has a minimum value of -8< p>

Cross point A to make the X-axis perpendicular line, and cross to C

AOB = 135°, then AOB = 45° in AOC

then |oc|=|ac|

And oc,ac is the horizontal and vertical coordinates, in this case, obviously, oc is a negative value, ac=1 2oc 2

oc|=|ac|

ac=2,oc=-2

Then the area of the triangle AOB = 1 2 * ob * ac = 1 2 * 3 * 2 = 3

When aob=135°, then the linear equation for oa is y=-x

Substituting y=x 2 2:

x=x^2/2

x^2+2x=0

x=0,x=-2.

So the a coordinate is (-2,2).

Then the area of the triangle AOB = 1 2*|ob|*|ya|=1/2*3*2=3

Because the original equation increases the function at once.

So the coefficient of the term is greater than one.

k2+3k+4)>0 to solve the fingers.

1. The primary function depends on its primary term coefficient, and the coefficient Daewoo 0 increases and decreases less than 0. If it is a second degree, it is necessary to determine whether it is a second degree function, and then search to determine whether the coefficient of the second order term is zero. If it is zero, it is done according to the missing number of the letter pants once, if not, it is necessary to look at the symmetry axis of the 2th function according to the 2nd method.

Let the side length of the subtracted square be x

Yes: (10-2x) * (8-2x) = 48

Yes: x 2-9x+8=0

Yes: (x-1)*(x-8)=0

Get: x1=1; x2 = 8 (rounded).

That is, the side length of the subtracted square is 1cm

Let the side length of the subtracted square be x

Side area s = 2x (10-2x) + 2x (8-2x) = 4 (9x - 2x 2).

Yes :s'=4(9-4x)

Ream'=0, get x=

That is, when x=, s has a maximum.

That is: s maximum = 4 (9*

Let the side length of the subtracted square be x, and when the top cover of the box of the cuboid is taken in the direction of 10cm, let the box length be b, and there are: 10=2x+2b

Yes: b=5-x

Yes: side area s=2x(8-2x)+2x(5-x)=2(13x-3x2).

Yes :s'=2(13-6x)

Ream'=0, which gives x=13 6

That is, when x=13 6cm, s has a maximum.

That is: s maximum = 2 * (13 6) * (13-3 * 13 6) = 169 6 (cm).

When the top cover of the box of the cuboid is taken in the direction of 8cm, the length of the box is a, and there is: 8=2x+2a

Yes: a=4-x

Yes: side area s=2x(10-2x)+2x(4-x)=2(14x-3x2).

Yes :s'=2(14-6x)

Ream'=0, which gives x=14 6

That is, when x = 14 6cm, s has a maximum.

That is: s maximum = 2 * (14 6) * (14-3 * 14 6) = 196 6

98/3 (cm²)

Comparing the two directions of taking the upper cover, it is obvious that when taking in the 8cm direction, the side area is larger, i.e., when x=7 3cm, there is a maximum side area of 98 3 (cm).

s1=2[(8-2x)x+x(10-2x)] to find the maximum value of this quadratic function.

s2=2[ x(8-2x)+x(10-2x) 2] s2=2[ x(8-2x) 2+x(10-2x)] Find the maximum of these two Take the largest one.

You're drawing a picture to see if my style is right? I can't guarantee it's right.

Let the side area be s and the side length of the square be x, then.

s=2x*(10-2x)+2x*(8-2x)=8(9/2x-x*x)

8[(81 16-(x-9 4)(x-9 4)]When x 9 4, s has a maximum value 8*81 16 81 2 The second question subtracts 2 squares of the same size and 2 squares of the same shape, respectively. A rectangle of the same size is a bit incomprehensible.

Solution: (1).

Let (x,y) be on f, then (-x,y) on e, and substitut e to y=x2-4x+3

This is the equation for f.

2) It is easy to find a(-3,0), b(-1,0), c(1,0), d(3,0), m(0,3), the parallelogram diagonals are bisected with each other, then the intersection point is the midpoint of cm (1 2, 3 2), then n is (4, 3), on f.

So there is n(4,3) that satisfies the topic. Thank you.

One. It is known that f(x) = x -12x+5

1. When x r, the minimum value of f(x) and the monotonic decrease interval and increase interval are found.

1,f(x)=x^2-12x+5

f'(x)=2x-12

Order f'(x)=0

x=6 then f(x) decrements on (-infinity, 6), takes the minimum value at x=6, and increments f(x)min=f(6)=-31 at (6,+infinity).

When x (0, closed 7), f(x) is the minimum and maximum values.

First, ask if the 0,7 in (0,7) is desirable, and if so, the following is preferable:

F(x) is obtained from 1 and takes the minimum value at 6, then f(x)min=-31 compares the size of f(0) and f(7), f(0)=5, and f(7)=-30 then f(0) is the largest.

Two. Knowing that the quadratic function f(x) passes through the point (0,5) and the axis of symmetry x=6 and the minimum value is -31, find the analytical domain of f(x).

Let's first ask if it is the analytical formula of f(x), if so, then:

Let f(x)=ax 2+bx+c

Because (0,5) is on the function, bringing the point into gives 5=c

Since the axis of symmetry is x=6, then -b 2a=6....1) The minimum value is -31, because the definition domain is not stated in the question, then it should be xr, and the function takes the minimum value at x=6.

Then -31 = 36a + 6b + 5....2)

(1) and (2).

a=1,b=-12

Three. Knowing that the quadratic function f(x) intersects with the x-axis at two intersections (1,0)(2,0) and with the y-axis at (0,2), the analytical formula of f(x) is obtained.

Let f(x)=ax 2+bx+c

Bring (1,0),(2,0),(0,2) into the function.

a=1,b=-3,c=2

f(x)=x^2-3x+2

Four. Knowing that f(x) has (6,0)(1,0)(2,0) to find f(0) analytical, "Are you sure you copied the wrong question!" ”

Establish. f(x)=ax 2+bx+c.

Bring in the function.

a=0,b=0,c=0

f(x)=0,f(0)=0

f(x)=x^2-12x+36-29（

Recipes) (x-6) 2-29

x-6) 2 is greater than or equal to 0

So. The minimum is 0

Monotonic subtraction interval.

infinity, 6) increase (6, + infinity esteem).

2. Max. Yes when x=0f(0)=

Utmost. When x=6

is -29f(x)=a(x-6) 2-31 by Foolish.

The axis of symmetry and the minimum accompaniment can be obtained.

And then. When x=0.

36a=36

a=1f(x)=(x-6)^2-31

Set the intersection type. f(x)=a(x-x1)(x-x2). a(x-1)(x-2)

is set to the general tense.

You're going to have to check it out.

Thanks for its contents.

The opening is downward, the axis of symmetry is to the right of the y-axis, and (0,1) and (-1,0) are a<0

b/2a > 0

c= 1a-b+c=0

From the first and second formulas, b>0 is obtained

then 2b>0

There is also a-b+c=0

Add it up to get a+b+c>0