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

Apr 14, 2018

$y = 8 x + 2$

#### Explanation:

Tangent Line to a Curve
simplify
$y = {x}^{3} - 2 {x}^{2} + x - 2$
differentiate
$y ' = 3 {x}^{2} - 4 x + 1$
substitute for $x = - 1$
$y ' = 8$
the value of $y '$ is the slope of the tangent line at $x = - 1$
and to get the point which lies on the tangent we substitute with $x = - 1$ in the function
$y = - 6$
so the point on the line is$\left(- 1 , - 6\right)$ and its slope is 8
substitute in the following formula to get the equation

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

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

$y = 8 x + 2$

Apr 14, 2018

$y = 8 x + 2$

#### Explanation:

At $x = - 1$:

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

$y = - 3 \times 2$

$y = - 6$

The tangent passes through $\left(- 1 , - 6\right)$

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

$y = {x}^{3} - 2 {x}^{2} + x - 2$

$y ' = 3 {x}^{2} - 4 x + 1$

At $x = - 1$:

$m = 3 {\left(- 1\right)}^{2} - 4 \left(- 1\right) + 1$

$m = 3 + 4 + 1$

$m = 8$

Since we have the gradient of the tangent and a point through which it passes we can use the point-gradient formula:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

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

$y = 8 x + 8 - 6$

$y = 8 x + 2$