Question

For `f(x) =3 x^2 + 6 x`, what is the equation of the line tangent to `x =13`?

Answer 1

In slope-intercept form, it is `y=84x-507`.

Explanation:

Step 1: Find the first derivative `f'(x)` using the power rule:

`d/dxax^n=nax^(n-1)`

So `f(x)=3x^2+6x`

`=>f'(x)=6x+6`

This gives us an equation to find the slope of `f(x)` at any given `x`-value.

Step 2: Find the slope at `x=13` by plugging this `x`-value into `f'(x).`

`f'(x)=6x+6`
`=>f'(13)=6(13)+6`
`color(white)(=>f'(13))=78+6`
`color(white)(=>f'(13))=84`

So when `x=13`, the slope is `m=f'(13)=84`.

Step 3: Find the `y`-coordinate for `x=13` using `y=f(x)`.

`f(x)=3x^2+6x`
`=>f(13)=3(13)^2+6(13)`
`color(white)(=>f(13))=3(169)+78`
`color(white)(=>f(13))=507+78`
`color(white)(=>f(13))=585`

So when `x=13`, we have `y=f(13)=585`.

Step 4: Plug these known values for `x`, `y`, and `m` into your favourite line equation. (I will use slope-point form: `y=m(x-x_1)+y_1`.)

`y=m(x-x_1)+y_1`
`=>y=84(x-13)+585`
`=>y=84x-1092+585`
`=>y=84x-507`

This is our equation for the line tangent to `f(x)` at `x=13`.

Note:

For consistency with calculus, the line equation may be written

`y=f'(x_0)(x-x_0)+f(x_0)`

The calculations are identical; the only difference is the symbols that are used as placeholders for the information we need:

`m=f'(x_0)`
`x_1=x_0`
`y_1=f(x_0)`