Find the slope of the tangent to the curve at point

Question

It should show each step

Answer 1

You haven't specified the equation of the curve or the point.

I can only show you the general method. Suppose y=f(x), then dy/dx=f'(x) (the derivative of the function with respect to x). So you must be able to differentiate the function.

dy/dx is the gradient or slope of the tangent.

dy/dx will be another function of x, which we can call g(x)=f'(x). The point (call it (a,b)) you are given has an x coordinate, x=a, as a number and the gradient=g(a), that is, substitute a for x in g(x).

The gradient will also be a number, which we can call "m".

The tangent is a straight line of slope m and we can write its equation:

y-b=m(x-a), y=mx-ma+b. The quantity b-ma is a number because we know a, b and m.

So here's another way to write the equation:

y=g(a)x-ag(a)+b or y=f'(a)x-af'(a)+b, where you need to substitute f'(a), a and b as actual numbers.

EXAMPLE

Let y=2x²+7x+8 be the equation of the curve so f(x)=2x²+7x+8, then dy/dx=f'(x)=4x+7.

Let the point be (1,17) (note that y=f(1)=17).

f'(1)=4+7=11, so m=11 and the equation of the tangent at the point is:

y-17=11(x-1), y=11x-11+17, y=11x+6.