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 p?

Let P(1,5) to be a point through f(x)=6x-x^2

-Let Q(x, 6x-x^2) be on the graph. Find slope of the secant line PQ for x=3, 2, 1.5, 1.01, 1.001.

-Use your aswer to guess the slope of the tangentto f(x) at P.

-Then find the equation of the tangent line at P.

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The line that passes through PQ is written applying the formula:

(xQ-xP)/(x-xP) = (yQ-yP)/(y-yP)

The slope of the line is m = (yQ-yP)/(xQ-xP)

y-yP = (yQ-yP)(x-xP)/(xQ-xP)

We'll substitute the coordinates of P and Q into the formula above:

y-5 = (6x-x^2-5)(x-1)/(x-1)

We'll find slope of the secant line PQ for x=3

m = (6*3-3^2-5)/(3-1)

m = (18 - 9 - 5)/2

m = 4/2

m = 2

We'll find slope of the secant line PQ for x = 2

m = (6*2-2^2-5)/(2-1)

m = (12-4-5)/1

m = 3

We'll find slope of the secant line PQ for x = 1.5

m = (6*1.5 - 1.5^2 - 5)/(1.5-1)

m = (9-2.25) / (0.5)

m = 1.75/0.5

m = 3.5

The tangent line to the function f(x), in the point P is the value of derivative of the function f(x), in that point.

We'll calculate the derivative of f(x).

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

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

f'(x) = 6 - 2x

f'(1) = 6 - 2*1

f'(1) = 4

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.

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