# How do you find the equation of a line tangent to the function y=x^2(x-2)^3 at x=1?

May 12, 2018

The equation is $y = 9 x - 10$.

#### Explanation:

To find the equation of a line, you need three pieces: the slope, an $x$ value of a point, and a $y$ value.

The first step is to find the derivative. This will give us important information about the slope of the tangent. We will use the chain rule to find the derivative.

$y = {x}^{2} {\left(x - 2\right)}^{3}$
$y = 3 {x}^{2} {\left(x - 2\right)}^{2} \left(1\right)$
$y = 3 {x}^{2} {\left(x - 2\right)}^{2}$

The derivative tells us the points what the slope of the original function looks like. We want to know the slope at this particular point, $x = 1$. Therefore, we simply plug this value into the derivative equation.

$y = 3 {\left(1\right)}^{2} {\left(1 - 2\right)}^{2}$
$y = 9 \left(1\right)$
$y = 9$

Now, we have a slope and an $x$ value. To determine the other value, we plug $x$ into the original function and solve for $y$.

$y = {1}^{2} {\left(1 - 2\right)}^{3}$
$y = 1 \left(- 1\right)$
$y = - 1$

Therefore, our slope is $9$ and our point is $\left(1 , - 1\right)$. We can use the formula for the equation of a line to get our answer.

$y = m x + b$

$m$ is the slope and $b$ is the vertical intercept. We can plug in the values we know and solve for the one we don't.

$- 1 = 9 \left(1\right) + b$
$- 1 = 9 + b$
$- 10 = b$

Finally, we can construct the equation of the tangent.

$y = 9 x - 10$