Question

Any and all help would be appreciated.
Find the equation of the tangent line to the curve y = (x - 2)^5 (x + 1)^2 at the point (3, 16)

Answer 1

The line passing through a point Pt(xt,yt) is
(y-yt) = m(x-xt)
where m is the slope.

For the given function
f(x) = (x - 2)^5 (x + 1)^2
We first validate that f(3) = 16, the tangent point.
f(3) = (3 - 2)^5 (3 + 1)^2 =16 ..... OK
The only unknown is the slope m, which we should be able to find by the derivative of the function, f'(x).

Differentiating f(x) as the product of two binomials, (x-2)⁵ and (x+1)², and utilising the product rule, d(uv)=udv+vdu, we get
f'(x) = (x-2)⁵*2(x+1) + 5(x-2)⁴(x+1)²

The slope m can be found readily by
m=y'(3)
=88
Thus the equation of the required tangent is:
(y-yt) = m(x-xt)
(y-16) = 88(x-3)
y=88x-248

Answer 2

Take the derivative dy/dx (that is the slope).

y'=5(x-2)^4(x+1)^2+2(x-2)^5((x+1)

Put in the point x=3

Now you have the slope at x=3

y=mx+b
you have y, m, x...solve for b.