Question

How do you find the tangent line to the curve `y=x^3-9x` at the point where `x=1`?

Answer 1

`y=-6x-2`

Explanation:

Given: `y=x^3-9x`.

Let `f(x)=y=x^3-9x`. At `x=1`, `f(x)=1^3-9*1=1-9=-8`.

So, the point we are targeting is `(1,-8)`.

To find the slope of the tangent line there, we must differentiate `f(x)` and then plug in `x=1` there.

`:.f'(x)=3x^2-9`

At `x=1`,

`f'(1)=3*1^2-9`

`=3-9`

`=-6`

So, the slope of the tangent line is `-6`.

Now, we use the point-slope formula to compute the equation, that is,

`y-y_0=m(x-x_0)`

• `(x_0,y_0)` are the original coordinates

Therefore, we get,

`y-(-8)=-6(x-1)`

`y+8=-6(x-1)`

`y+8=-6x+6`

`y=-6x+6-8`

`=-6x-2`

A graph shows it:

Answer 2

`y=-6x-2`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = 1"`

`rArrdy/dx=3x^2-9`

`rArrdy/dx(x=1)=3-9=-6`

`rArry(x=1)=1-9=-8rArr(1,-8)`

`"using "m=-6" and "(x_1,y_1)=(1,-8)`

`y+8=-6(x-1)`

`rArry=-6x-2larrcolor(red)"equation of tangent"`