Use the function y=x^3-2x^2-3x+4, approximate the slope to the curve at x=2
y'(x) = 3x^2-4x-3
y'(2) = 12-8-3 = 1
But that is exactly the slope of the curve at x=2. Why bother to approximate? I suspect something is missing.
No I was just making sure. Thanks
To approximate the slope to the curve at a particular point on the graph of a function, we can use the concept of the derivative. The derivative of a function at a given point represents the slope of the tangent line to the graph at that point.
Step 1: Find the derivative of the function y = x^3 - 2x^2 - 3x + 4.
To find the derivative, we can apply the power rule:
- For terms involving x raised to a power, we multiply the term by the power and then decrease the power by 1.
So, dy/dx = 3x^2 - 4x - 3.
Step 2: Substitute the given x-value into the derivative expression.
Plug x = 2 into the derived expression: dy/dx = 3(2)^2 - 4(2) - 3.
This simplifies to dy/dx = 12 - 8 - 3 = 1.
Therefore, the slope of the curve at x = 2 is approximately 1.