Find slope of the tangent to f(x) at P and then find equation of the tangent line at P. Is the derivative of the equation of the tangent line at...

f(x) = 6x-x^2.

P(1,5) is
point on the curve as f(1) = 6*1*1^2 = 5 verifies.

The
slope of the curve at (1,5) is:

dy/dx = (6x-x^2)' =
6-2x.

So (dy/dx at x= 1 ) is 6-2*1 =
4.

So the equation of tangent at (x1,y1) is y-t1 =
(dy/dx)(x-x1)

But (x1,y1) = (1,5) and (dy/dx at x = 1) =
4.

So the equation of tangent at (1,5)
is:

y-5 = 4(x-1) .

4x-y -4+5 =
0

4x-y+1 = 0.

The slope secant
value  PQ :

slope = (Py- Qy)/(Px-
Qx).

Qx = 3, 2, 1.5,   1.01,
1.001

Qy =9,  8,  6.75,  5.0399, 
5.00399.

The value of the slope
:

Slope : 
(Py-Qy)/(Px-Qx)

(5-9)/(1-3) =
2

(5-8)/(1-2)
=3

(5-6.75)/(1-1.5) =
3.5

(5-5.0399)/(1-1.01) =
3.99

(5-5.003999)/(1-1.001) =
3,999

The process shows that limit of the secant PQ as Q
tends P is dy/dx = 4.