Two quadratic function application questions, quadratic function application questions

1: The output of X-file products is: 76-(X-1)*4, and the profit per piece is: 10+(X-1)*2

y=(76-4x+4)(10+2x-2)=(80-4x)(8+2x)=640+160x-32x-8x^2

y=-8x 2+128x+640 (1=When y=1080, 1080=-8x 2+128x+640.)

8x^2-128x+440=0

x^2-16x+55=0

x-11)(x-5)=0

x=11 (rounded), x=5

Therefore, it is for the production of 5th gear products.

2.Vedic theorem, x1+x2=-b a=b c=2 3

x1x2=c/a=a/c=-1/3

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

Function analytical: y=-3ax2-2ax+a

Parabolic analytical: y=ax2-2ax-3a=a(x-3)(x+1).

a(-1,0),b(3,0),p(1,-4a)

kap=-4a/2=-2a,kbp=-4a/-2=2a

kap*kbp=-1

4a2=1a=1/2,-1/2

Function analytical: y=-3x 2 2-x+1 2

or, y=3x2 2+x-1 2

Parabolic analytical: y=(x-3)(x+1) 2 or y=-(x-3)(x+1) 2

Second question c(0,a).

tancab=a,tancba=a/3

a*a/3=a^2/3=1,a=√3，-√3

Parabolic analytical: y=(x-3)(x+1) 3, or, y=-(x-3)(x+1) 3

1)y=[76-4*(x-1)]*10+2(x-1)]=(80-4x)(2x+8)=-8(x-20)(x+4)

8(x^2-16x^2-80)

2)y=1080=-8(x^2-16x^2-80)

x=5 2, the two roots of the equation cx -bx+a=0 are 1 3 and 1

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

y=-1/3cx^2+2/3cx+c，xa+xb=2，xa*xb=-3

The coordinates of a,b are (-1,0), (3,0), and the vertices are.

1,(-4 3c 2-4 9c 2) -4 3c)=(1,4 3c)=p,is a right angle So ap*bp=0,(2,4 3c)(-2,4 3c)=0,c=plus or minus (root number 6) 2

Substitute C for you.

tan cab·tan cab=1, that is, the angle abc = 90 degrees, the first question has been obtained, you can use p with c.

I don't know, but you can check it out on the education network.

(1) Profit per loaf: x-5; Number of sells: 160-(x-7)20=300-20x

2) Set: y=ax 2+bx+c

y=(300-20x)(x-5)

20x^2+400x-1500

3) Let y=a(x-h) 2-k

y=-20(x^2-20x+100)-1500+2000=-20(x-10)^2+500

When x=10, y is the maximum, and the maximum value is 500

Answer: When the price of bread is 10 jiao, one yuan, the maximum profit is 50 yuan.

1. Profit per loaf: x-5

Number of units sold: 300-20x

2、y=(160-20(x-7))(x-5）=-20x²+400x-1500

3、y=-20x²+400x-1500

20(x-10)²+500

When x=10, y has a maximum value of 500 angles.

(1) Profit per loaf: x-5;

Number of loaves sold: 160-20 (x-7).

2)y=(x-5)[160-20(x-7)]=(x-5)(300-20x)

300x-20x²-1500+100x

20(x²-20x+75)

20(x-10)²+500

3) When the unit price of bread is 10 jiao, the store makes the most profit by selling this kind of flour every day. The maximum profit is 500 jiao.

Solution: (1) This is obviously a piecewise function, y=20- (x-100) 10

100 x 200, when x=200 yuan, y=28-16=12 (10,000 pieces).

y=12- (x-200) 10 1=,200 x 300,2) The investment cost is 480 + 1520 = 20 million yuan.

y=，100≤x＜200，w=xy-40y-2000

x-40）（

It can be seen that in the first year, 100 x 200 is destined to lose, and the loss is the least at x = 195, which is 780,000 yuan.

200≤x≤300，y=，w=xy-40y-2000

x-40）（

It can be seen that the first year is destined to lose money at 200 x 300, and the loss is the least when x = 200, which is 800,000 yuan.

To sum up, the loss at x=195 is the least, which is 780,000 yuan

3) The total profit of two years is not less than 1842 yuan, it can be seen that the second year should be at least 1842 + 78 = 19.2 million yuan, since the two years are counted together, we do not need to calculate the investment cost of 20 million yuan in the second year

Year 2: 100 x 200 hours.

Profit in the second year = xy-40y=

Solving inequality yields: 190 x 200

200 x 300.

Profit in the second year = xy-40y=

Solving the inequality yields: 160 x 200, combined 200 x 300, which is only x=200

In summary, 190 x 200 is the solution.

At this time, if we look at y=, we can see that when x=190, y is the largest, which is.

So when the price is 190 yuan, the sales volume is the largest

Xie Conghe: Set a price reduction for each piece and shout about x yuan.

20+2x)(40-x)=1200

40-x)(x+10)=600

x^2-30x+200=0

x-10)(x-20)=0

x=10 or x=20

Answer: Each piece is reduced by 10 yuan or 20 yuan, and you can make a profit of 1200 yuan every day.

y=(20+2x)(40-x)

2x^2+60x+800

y=-2(x^2-30x+225)+450+800-2(x-15)^2+1250

When x = 15, y has a maximum value of 1250

Answer: When the price is reduced by 15 yuan, the daily profit seepage is the largest, which is 1250 yuan.

y=t(x-42)=-3x^2+204x+126x-204*42=-3x^2+330x-8586=-3(x^2-110x+3025)+407=-3(x-55)^2+507

If other factors are not considered, if the mall wants to get the maximum sales profit every day for selling this batch of clothing, the sales profit per piece should be set at 55 yuan, and the maximum sales profit per day is 507

An engineering team contracted a demolition project in the implementation of shantytown reconstruction in our city. The original plan was to demolish 1250 per day, because of the lack of preparation, the first day of the demolition of 20 less, from the second day, the project to speed up the demolition, the third day of demolition 1440

Seeking: (1) The area of the engineering team on the first day of demolition?

Solution: 1250 (1-20%) = 1000 square meters on the first day of demolition.

2) If the demolition area of the second and third days of the project is the same percentage as the increase in the hidden land on the previous day, the filial piety file is asked for this percentage.

Let this percentage be a

1000×（1+a）²=1440

1+a）²=

1+a=a=20% is 20%.

Solution: (1) Set the maximum to n, answer: customers can buy at least 50 at a time to buy at the lowest price (2) y=[

+9*x squared

+ A: --

3) When x=45, y gets the maximum value.

So the lowest price =

A: --

Solution: (1)20, a=50 (pcs) 3 points (2) y=4x or y=-1 10x square + 9x (3) because you sell more and earn more, that is, y increases with the increase of x 8 points It can be seen from the quadratic function image, when y is the largest, 9 minutes at this time, the lowest selling price is 20 yuan) 10 points.

(1) The merchant makes a profit of 4 yuan, and the lowest price purchase is not profitable, that is, the purchase price **, 4, indicating that you want to buy 40 more, plus the first 10, a total of 50.

2) The current price is 20-(x-10)*, and the number of units is x, so y=[20-(x-10)*

3) Let the number be increased to a, (16=10+(20-a), y1=(20-x)*x, x<10+(20-a), y2=[20-(x-10)*, ensure y1>y2, and solve the x range. And then find A

（1）y=500-（x-50）*10

y=1000-10x 50<>75

The sales price should be set at $80.