Question

What is the equation of the line tangent to `f(x)=x(x-3)^2` at `x=2`?

Answer 1

`y=-3x+8`

Explanation:

Find the point the tangent line will intercept:

`f(2)=2(2-3)^2=2(-1)^2=2`

The tangent line will pass through the point `(2,2)`.

To find the slope of the tangent line, find the value of the derivative at `x=2`.

To find `f'(x)`, we will use the product rule.

`f'(x)=(x-3)^2d/dx[x]+xd/dx[(x-3)^2]`

Find the value of each derivative. The second will require a simple application of the chain rule. Alternatively, you could say that `(x-3)^2=x^2-6x+9`, which makes the derivative findable through the power rule.

`f'(x)=(x-3)^2(1)+x(2(x-3))`

This can be simplified.

`f'(x)=x^2-6x+9+2x^2-6x`

`f'(x)=3x^2-12x+9`

Note that the derivative also could have been found through distribution of `f(x)`, as follows:

`f(x)=x(x^2-6x+9)`

`f(x)=x^3-6x^2+9x`

`f'(x)=3x^2-12x+9`

Regardless of how you got there, the derivative is the same. Now that we have the derivative, we can find the slope of the tangent line:

`f'(2)=3(2^2)-12(2)+9=12-24+9=-3`

The tangent line passes through the point `(2,2)` and has slope `-3`, giving the linear equation

`y-2=-3(x-2)`

or

`y=-3x+8`

Graphed are the original function and its tangent line:

graph{(x(x-3)^2-y)(y+3x-8)=0 [-1, 5, -2.526, 6.364]}

Answer 2

y + 3x - 8 = 0

Explanation:

The equation of the tangent is : y - b = m(x - a )

where m= gradient and (a , b ) a point on the line.

m and (a , b ) are required to be found. The gradient of tangent

is f'(x) and a = 2 . To find b substitute x = 2 into f(x).

Differentiate using `color(blue)(" product and chain rules ")`

f'(x) `= x d/dx(x-3)^2 + (x-3)^2 d/dx (x)`

`= x [2(x-3) d/dx(x-3)] + (x-3)^2 .1`

`= 2x(x-3) .1 + (x-3)^2` = (x-3)(2x + x - 3 )

= 3 (x-3)(x-1)

now m = f'(2) = 3(2-3)(2-1) = -3

and b = f(2) =`2(-1)^2 = 2`

equation is : y-2 = -3(x-2)

so y - 2 = -3x + 6

hence y + 3x - 8 = 0