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

Apr 21, 2018

$y = - 6 x - 2$

#### Explanation:

Given: $y = {x}^{3} - 9 x$.

Let $f \left(x\right) = y = {x}^{3} - 9 x$. At $x = 1$, $f \left(x\right) = {1}^{3} - 9 \cdot 1 = 1 - 9 = - 8$.

So, the point we are targeting is $\left(1 , - 8\right)$.

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

$\therefore f ' \left(x\right) = 3 {x}^{2} - 9$

At $x = 1$,

$f ' \left(1\right) = 3 \cdot {1}^{2} - 9$

$= 3 - 9$

$= - 6$

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

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

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

• $\left({x}_{0} , {y}_{0}\right)$ are the original coordinates

Therefore, we get,

$y - \left(- 8\right) = - 6 \left(x - 1\right)$

$y + 8 = - 6 \left(x - 1\right)$

$y + 8 = - 6 x + 6$

$y = - 6 x + 6 - 8$

$= - 6 x - 2$

A graph shows it:

Apr 21, 2018

$y = - 6 x - 2$

#### Explanation:

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 9$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(x = 1\right) = 3 - 9 = - 6$

$\Rightarrow y \left(x = 1\right) = 1 - 9 = - 8 \Rightarrow \left(1 , - 8\right)$

$\text{using "m=-6" and } \left({x}_{1} , {y}_{1}\right) = \left(1 , - 8\right)$

$y + 8 = - 6 \left(x - 1\right)$

$\Rightarrow y = - 6 x - 2 \leftarrow \textcolor{red}{\text{equation of tangent}}$