# What is the equation of the line that is normal to f(x)=-2x^2+4x-2  at  x=3 ?

Feb 16, 2016

$y = \frac{1}{8} x - \frac{67}{8}$

#### Explanation:

To find the normal line's slope, first find the slope of the tangent line at the same point. You can find the tangent line's slope by finding the value of the derivative at $x = 3$.

The derivative of the function can be found through the power rule:

$f \left(x\right) = - 2 {x}^{2} + 4 x - 2$
$f ' \left(x\right) = - 4 x + 4$

The slope of the tangent line is

$f ' \left(3\right) = - 4 \left(3\right) + 4 = - 8$

Now, to find the slope of the normal line, take the opposite reciprocal of $- 8$, since the normal line and tangent line are perpendicular so their slopes are opposite reciprocals.

The opposite reciprocal of $- 8$ and slope of the normal line is $1 \text{/} 8$.

The normal line intercepts the function at the point $\left(3 , f \left(3\right)\right) = \left(3 , - 8\right)$.

(Don't be confused by the fact that both $f \left(3\right) = - 8$ and $f ' \left(3\right) = - 8$. This is the product of pure chance.)

The equation of the normal line can be written in point-slope form:

$y + 8 = \frac{1}{8} \left(x - 3\right)$

In slope-intercept form, this is

$y = \frac{1}{8} x - \frac{67}{8}$

Graphed are the function and its normal line:

graph{(y+2x^2-4x+2)(y-x/8+67/8)((x-3)^2+(y+8)^2-.1)=0 [-15.11, 20.93, -15.42, 2.59]}