Question

Find all tangent lines to the graph of `y = x^3` that pass through the point. `P(2, 4)`?

Answer 1

`{(y-4=(12-6sqrt3)(x-2)),(y-4=3(x-2)),(y-4=(12+6sqrt3)(x-2)):}`

Explanation:

We know two points on the tangent line: the given point `P(2,4)` and the point of tangency located on the graph of `y=x^3`: the point `P(x,x^3)`.

We know that the slope of the tangent line will be equal to the value of the derivative of `y` at the intended value.

The derivative of `y=x^3` is `dy/dx=3x^2`.

Thus, `3x^2` will be equal to the slope of the line passing through the points `P(2,4)` and `P(x,x^3)`.

Setting this up yields:

`3x^2=(x^3-4)/(x-2)`

Now solving:

`3x^2(x-2)=x^3-4`

`3x^3-6x^2=x^3-4`

`2x^3-6x^2+4=0`

To solve this, note that `2-6+4=0`, so `x=1` is a root and `x-1` is a factor of the polynomial.

Do the long division of `(2x^3-6x^2+4)/(x-1)` to find the remaining quadratic factor:

`(x-1)(2x^2-4x-4)=0`

`2(x-1)(x^2-2x-2)=0`

Note that `x^2-2x-2=0` gives solutions of `1+-sqrt3`.

We now know the `x` values for the points of tangency where the tangent lines pass through `P(2,4)`: `1-sqrt3,1,` and `1+sqrt3`.

Find the slopes of the tangent lines by finding the value of the derivative at each of these points:

`f'(1-sqrt3)=3(1-sqrt3)^2=3(1-2sqrt3+3)=12-6sqrt3`

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

`f'(1+sqrt3)=3(1+sqrt3)^2=3(1+2sqrt3+3)=12+6sqrt3`

Then, writing the equations of the three lines with respective slopes of `12-6sqrt3,3,` and `12+6sqrt3` passing through the point `P(2,4)`, we obtain the tangent lines:

`{(y-4=(12-6sqrt3)(x-2)),(y-4=3(x-2)),(y-4=(12+6sqrt3)(x-2)):}`