Question

What is the equation of the tangent line of `f(x)=(x-1)^3+(x-2)(x-7)` at `x=7`?

Answer 1

`y=113x-545`

Explanation:

Step 1: Determine the y coordinate at the point `x=7`

`f(x)=(x-1)^3+(x-2)(x-7)`
`f(7)=(7-1)^3+(7-2)(7-7)rArr6^3+(5*0)rArr216`
Therefore the tangent to the curve passes through the point `color(orange)((7;216))`

Step 2: Expand and simplify the given Function

`f(x)=(x-1)^3+(x-2)(x-7)`
`rArrx^3-3x^2+3x-1+x^2-7x-2x+14`
`rArrx^3-2x^2-6x+13`

Step 3: Find the Derivative

`f'(x)=3x^2-2*(2x)-6`
`rArr3x^2-4x-6`

Step 4: Calculate the `color(brown)(Gradien)t` of the `color(green)("Tangent")`

substituting `x=7` into the equation `f'(x)` :
`f'(7)=(3*7^2)-(4*7)-6rArr113`

As we have the `color(brown)(Gradien)t` of the `color(green)("Tangent")` we can write that the equation of `color(green)("Tangent")` is

`y=113x+c`

where c is the `y`-intercept

Step 4: Determine the value of c

substituting `x=7` & `y=216` into the above equation:
`216=(113*7)+c`
`:.c=-545`

Step 5: Determine the equation of Tangent

Now, substituting the value of `c=-545`

`color(red)(y=113x-545)`
graph{(x^3-2x^2-6x+13-y)(113x-545-y)=0 [-10, 10, -5, 5]}