Math Problem Statement

Find the expression of a quadratic function f(x) = ax^2 + bx + c, where the function passes through points A(0, 1) and B(2, 5), and the line is tangent to the curve at A(0, 1).

Solution

This task involves finding the expression of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c that satisfies certain conditions given in the problem. Let’s analyze the provided details:


Problem Setup

  1. The function ff is a quadratic polynomial defined by: f(x)=ax2+bx+cf(x) = ax^2 + bx + c where a,b,ca, b, c are real coefficients.

  2. Two specific points are given:

    • A(0,1)A(0, 1): The function passes through this point. f(0)=1f(0) = 1
    • B(2,5)B(2, 5): The function also passes through this point. f(2)=5f(2) = 5
  3. The line dd is tangent to the curve C\mathcal{C} at the point A(0,1)A(0, 1).


Strategy for Solution

Step 1: Use the given points to generate two equations.

  • Since f(0)=1f(0) = 1, substituting into the general form: f(0)=a(0)2+b(0)+c=c=1f(0) = a(0)^2 + b(0) + c = c = 1 Thus, c=1c = 1.

  • Now, use the second point B(2,5)B(2, 5): f(2)=a(2)2+b(2)+1=5f(2) = a(2)^2 + b(2) + 1 = 5 Simplifying: 4a+2b+1=54a+2b=4(1)4a + 2b + 1 = 5 \quad \Rightarrow \quad 4a + 2b = 4 \quad (1)

Step 2: Find the derivative of f(x)f(x) and apply the tangent condition.

The derivative of the function f(x)=ax2+bx+cf(x) = ax^2 + bx + c is: f(x)=2ax+bf'(x) = 2ax + b

  • At point A(0,1)A(0, 1), the slope of the tangent line is given by: f(0)=2a(0)+b=bf'(0) = 2a(0) + b = b This slope bb is the same as the slope of the line dd, which passes through points A(0,1)A(0, 1) and B(2,5)B(2, 5).

  • The slope between A(0,1)A(0, 1) and B(2,5)B(2, 5) is: slope=5120=42=2\text{slope} = \frac{5 - 1}{2 - 0} = \frac{4}{2} = 2 Thus, b=2b = 2.

Step 3: Solve for aa using the equations.

Now that we know b=2b = 2, substitute it into equation (1): 4a+2(2)=44a + 2(2) = 4 4a+4=44a=0a=04a + 4 = 4 \quad \Rightarrow \quad 4a = 0 \quad \Rightarrow \quad a = 0


Final Expression of the Function

Thus, the quadratic function is: f(x)=0x2+2x+1=2x+1f(x) = 0 \cdot x^2 + 2x + 1 = 2x + 1

This means that the quadratic function simplifies to a linear function f(x)=2x+1f(x) = 2x + 1, as the coefficient aa is zero.


Do you have any further questions about this solution, or would you like additional details?


Related Questions

  1. What is the general equation of the tangent line to a quadratic function at a given point?
  2. How does the slope of a line relate to the derivative of a function?
  3. Can a quadratic function ever reduce to a linear function? If so, under what conditions?
  4. What methods can be used to solve systems of equations involving multiple variables?
  5. What are some applications of tangents in optimization problems?

Tip: When working with quadratic functions, always check if the function reduces to a simpler form, like a linear function, under specific conditions.

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Math Problem Analysis

Mathematical Concepts

Quadratic functions
Tangents
Algebra

Formulas

Quadratic equation: f(x) = ax^2 + bx + c
Tangent condition: f'(x) = 2ax + b
Slope formula: m = (y2 - y1) / (x2 - x1)

Theorems

Derivative as the slope of the tangent
System of equations for determining unknown coefficients

Suitable Grade Level

Grades 10-12