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

Jul 13, 2016

The equation of the line will be $y = \frac{1}{9} x + \frac{137}{9}$.

#### Explanation:

Tangent is when the derivative is zero. That is $4 x - 1 = 0. x = \frac{1}{4}$ At x = -2, f' = -9, so the slope of the normal is 1/9. Since the line goes through $x = - 2$ its equation is $y = - \frac{1}{9} x + \frac{2}{9}$

First we need to know the value of the function at $x = - 2$
$f \left(- 2\right) = 2 \cdot 4 + 2 + 5 = 15$

So our point of interest is $\left(- 2 , 15\right)$.
Now we need to know the derivative of the function:
$f ' \left(x\right) = 4 x - 1$

And finally we'll need the value of the derivative at $x = - 2$:
$f ' \left(- 2\right) = - 9$

The number $- 9$ would be the slope of the line tangent (that is, parallel) to the curve at the point $\left(- 2 , 15\right)$. We need the line perpendicular (normal) to that line. A perpendicular line will a negative reciprocal slope. If ${m}_{| |}$ is the slope parallel to the function, then the slope normal to the function $m$ will be:
$m = - \frac{1}{{m}_{| |}}$

This means the slope of our line will be $\frac{1}{9}$. Knowing this we can proceed with solving for our line. We know it will be of the form $y = m x + b$ and will pass through $\left(- 2 , 15\right)$, so:
$15 = \left(\frac{1}{9}\right) \left(- 2\right) + b$
$15 + \frac{2}{9} = b$
$\left(\frac{135}{9}\right) + \frac{2}{9} = b$
$b = \frac{137}{9}$

This means our line has the equation:
$y = \frac{1}{9} x + \frac{137}{9}$