Math Problem Statement
Solution
It looks like you've uploaded an image from a worksheet titled "Maxima and Minima of Quadratic Functions 1". The first question from the image reads:
1. Given the quadratic function , the maximum value is 4 at , and the graph passes through the point . Find the values of the constants , , and .
To solve this:
Step 1: Use the Maximum Condition
The maximum value occurs at , so:
- At , . Substituting this into the quadratic equation: This simplifies to:
Step 2: Use the Point on the Graph
The quadratic also passes through the point , meaning:
- At , . Substituting into the quadratic equation: This simplifies to:
Step 3: Use the Maximum Slope Condition
At the maximum point, the derivative since the slope is zero. The derivative of is: At , the slope is zero, so: This simplifies to:
Step 4: Solve the System of Equations
Now we have three equations:
We can solve this system of equations to find , , and .
Would you like me to continue with solving the system, or do you have any specific part you'd like to focus on?
5 Related Questions:
- How do we solve a system of equations with substitution or elimination?
- What is the significance of the derivative being zero at a maximum or minimum point?
- How does the shape of a parabola relate to its coefficient ?
- Can a quadratic function have both a maximum and minimum value?
- How do changes in the constants and affect the position of a parabola?
Tip:
Always check the condition of the slope being zero at maxima or minima when dealing with quadratic functions!
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Math Problem Analysis
Mathematical Concepts
Quadratic Functions
Maxima and Minima
Differentiation
Formulas
Quadratic function: y = ax^2 + bx + c
Derivative: y' = 2ax + b
Maximum value condition: y'(x) = 0
Theorems
Derivative Test for Maxima and Minima
System of Linear Equations
Suitable Grade Level
Grade 9-11
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