Finding the Values of a and b for the Curve (y 2x^2 ax b) with Given Gradient

Differentiating a curve and setting its derivative to a given gradient can help us find the unknown constants in the curve's equation. In this guide, we'll walk through an example to find the values of a and b for the curve y 2x^2 ax b at the point (2, 11) where the gradient is 20.

Derivative and Gradient

The gradient of the curve at a specific point is given by the first derivative of the curve's equation evaluated at that point. In this case, the given gradient at the point (2, 11) is 20. This can be expressed as:

Gradient: y' 20

Step 1: Differentiate the Curve Equation

We start by differentiating the curve's equation y 2x^2 ax b with respect to x:

y' 4x a

Step 2: Use the Given Gradient

Since we know the gradient at the point (2, 11) is 20, we can substitute x 2 and y' 20 into the derivative to find a:

When x 2, y' 4(2) a 20

8 a 20

a 12

Step 3: Substitute a into the Original Equation

With a 12, we substitute it back into the original curve equation to find b. We use the point (2, 11) to do this:

11 2(2^2) 12(2) b

11 2(4) 24 b

11 8 24 b

11 32 b

b 11 - 32

b -21

Conclusion

The values of a and b for the curve y 2x^2 ax b at the point (2, 11) where the gradient is 20 are:

The final equation of the curve is:

y 2x^2 12x - 21

Further Explanation and Graphical Representation

The derivative y' 4x 12 gives us the slope at any point on the curve. By setting y' 20, we find the x-coordinate where the gradient is 20, which is x 2. This is confirmed by the original point (2, 11).

The graphical representation of the curve (y 2x^2 12x - 21) and its tangent line at (2, 11) with the slope 20 provides a visual aid to understand the problem better.