Solving a Parabolic Equation Given Specific Points and Gradient

In this article, we will explore how to determine the coefficients of a quadratic equation given specific points and the gradient at a particular point. This involves solving a system of equations and understanding derivatives to find the desired values.

Problem Statement

The task is to find the values of (a), (b), and (c) for the quadratic equation (y ax^2 bx c). We are given two key pieces of information: the curve passes through the point ((-1, 4)) and the gradient at the point ((2, 7)) is 7.

Step-by-Step Solution

Let's break down the problem into steps and solve it systematically.

Step 1: Use the Point (-1, 4)

Substituting the point ((-1, 4)) into the equation:

[4 a(-1)^2 b(-1) c]

This simplifies to:

[4 a - b c quad text{(Equation 1)}]

Step 2: Use the Point (2, 7)

Substituting the point ((2, 7)) into the equation:

[7 a(2)^2 b(2) c]

This simplifies to:

[7 4a 2b c quad text{(Equation 2)}]

Step 3: Use the Gradient at (2, 7)

The gradient of the curve (y ax^2 bx c) is given by the derivative:

[frac{dy}{dx} 2ax b]

At (x 2), the gradient is 7:

[7 2a(2) b]

This simplifies to:

[7 4a b quad text{(Equation 3)}]

Solving the Equations

We now have three equations:

(4 a - b c quad text{(Equation 1)} (7 4a 2b c quad text{(Equation 2)} (7 4a b quad text{(Equation 3)}

Let's solve these equations step by step.

Step 4: Express (b) in Terms of (a)

From Equation 3:

[b 7 - 4a]

Step 5: Substitute (b) into Equation 1

Substituting (b 7 - 4a) into Equation 1:

[4 a - (7 - 4a) c]

This simplifies to:

[4 a - 7 4a c]

[4 5a - 7 c]

[c 11 - 5a quad text{(Equation 4)}]

Step 6: Substitute (b) into Equation 2

Substituting (b 7 - 4a) into Equation 2:

[7 4a 2(7 - 4a) c]

This simplifies to:

[7 4a 14 - 8a c]

[7 -4a 14 c]

[c -4a 7 quad text{(Equation 5)}]

Step 7: Set Equations 4 and 5 Equal to Each Other

[11 - 5a -4a 7]

[11 - 7 -4a 5a]

[4 a]

[a 2]

Step 8: Find (b) and (c)

Using (a 2) in Equation 3:

[b 7 - 4(2) 7 - 8 -1]

Using (a 2) in Equation 4:

[c 11 - 5(2) 11 - 10 1]

Conclusion

The values are:

[a 2, quad b -1, quad c 1]

Thus, the equation of the curve is:

[y 2x^2 - x 1]

Mathematical Foundation and Relevance

This problem involves the application of quadratic equations, derivatives, and algebraic manipulation. It is relevant in various fields including engineering, physics, and economics, where the behavior of parabolic curves is often modeled and analyzed.