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Authoritative PPT Summary
"Application of Quadratic Functions" PPT Courseware 2
learning target
1. Be able to analyze and represent the quadratic function relationship between variables in practical problems under different backgrounds, and be able to use the knowledge of quadratic functions to solve the maximum or minimum value problems in practical problems.
2. Experience the process of exploring the maximum or minimum area of a rectangle, further gain experience in using mathematical methods to solve practical problems, and feel the application value of mathematical model ideas and mathematics.
3. Through the analysis of specific examples in life, understand mathematics in life and cultivate a love for mathematics.
Use a fence to surround a rectangular vegetable garden with one side against the wall. It is known that the length of the fence is 60m. Question: How should it be designed to maximize the area of the vegetable garden? What is the maximum area?
Solution: As shown in the figure, assuming the width of the rectangular vegetable garden is x(m), then the length of the vegetable garden is (60-2x)m and the area is y(�).
According to the meaning of the question, the functional expression between y and x is
y=x(60-2x)
=-2(x²-30x)
=-2(x²-30x+225-225)
=-2[(x-15)²-225]
=-2(x-15)²+450
Because a<0, the parabola opens downward, and the vertex (15,450) is the highest point of the image. When x=15, y has a maximum value, and the maximum value is 450. From the meaning of the question, we can know: 0
Therefore, when the width of the vegetable garden is 15m, the vegetable garden has the largest area, and the maximum area is 450 �.
Timely summary
Generally, because the vertex of the parabola y=ax²+bx+c is the lowest (high) point of the parabola, when x=-b/2a, the quadratic function y=ax²+bx+c has the smallest (largest) value, The minimum (maximum) value is 4ac-b²/4a.
Life is with me
As shown in the figure, ABCD is a square iron plate with a side length of 2m. Take a point M on the side AB and intercept two adjacent square plates with AM and MB as the sides. When the length of AM is what is the length of the intercepted plate? The smallest area?
Analysis: The area of the intercepted plate = the area of the square AMPQ + the area of the square MBEF. From what is known, a quadratic function can be constructed, and the properties of the quadratic function can be used to solve...
General steps for solving function word problems:
Assume unknowns (determine independent variables and functions);
Find equivalence relations and list functional expressions;
Simplify and organize into standard forms (linear functions, quadratic functions, etc.);
Find the value range of the independent variable;
Use function knowledge to solve (usually the optimal value problem);
Write the answer.
Homework questions
1. The shape of the window is a rectangle with a semicircle on top. The perimeter of the window is equal to 6m. How should the size of the window be designed so that it can transmit the most light?
2. As shown in the picture, a circular fountain is to be built in the park, and a pillar OA is installed in the center of the pool perpendicular to the water surface. O is exactly in the center of the water surface, OA=1.25m. Water is sprayed outward from the nozzle at A at the top of the column. The water flow falls along a parabola with the same shape in all directions. In order to make the water flow shape more beautiful, it is required to design the water flow to reach a maximum height of 2.25 meters from the water surface at a distance of 1 meter from OA. (1) If other factors are not considered, what is the minimum radius of the pool so that the sprayed water will not fall outside the pool? (2) If the parabolic shape of the water jet is the same as
(1) Same. The radius of the pool is 3.5 meters. To prevent the water flow from falling outside the pool, how many meters should the maximum height of the water flow reach?
(accurate to 0.1 meter)
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