What is the equation of the line normal to f(x)=(x+1)^2-4x+5 at x=-2?

Jul 11, 2018

$6 y - x - 86 = 0$

Explanation:

$f \left(x\right) = {\left(x + 1\right)}^{2} - 4 x + 5$
$f \left(x\right) = {x}^{2} + 2 x + 1 - 4 x + 5$
$f \left(x\right) = {x}^{2} - 2 x + 6$
When $x = - 2$, $f \left(- 2\right) = {\left(- 2\right)}^{2} - 2 \left(- 2\right) + 6 = 14$

$f ' \left(x\right) = 2 x - 2$

To find the gradient of $f \left(x\right)$ when $x = - 2$, you sub $x = - 2$ into $f ' \left(x\right)$
$f ' \left(- 2\right) = 2 \left(- 2\right) - 2$
$f ' \left(- 2\right) = - 4 - 2$
$f ' \left(- 2\right) = - 6$

What you just found is the gradient of the tangent of the line. Since you want to find the gradient of the normal of the line, we can use the formula ${m}_{1} {m}_{2} = - 1$ where ${m}_{1}$ and ${m}_{2}$ are the gradients of two different lines. We already know that one of the lines has a gradient equal to $- 6$ so that means the other line has a gradient equal to $\frac{1}{6}$

Equation of the normal
$\left(y - 14\right) = \frac{1}{6} \left(x + 2\right)$
$y - 14 = \frac{1}{6} x + \frac{1}{3}$
$6 y - 84 = x + 2$
$6 y - x - 86 = 0$