What is the equation of the line normal to  f(x)=-(x-1)(x+3)+4x^2-8x+2 at  x=0?

Feb 28, 2016

$y = \frac{1}{10} x + 5$

Explanation:

First simplify the function.

$f \left(x\right) = - \left({x}^{2} + 2 x - 3\right) + 4 {x}^{2} - 8 x + 2$

$f \left(x\right) = 3 {x}^{2} - 10 x + 5$

Before finding the normal line at $x = 0$, find the point the normal line will intercept:

$f \left(0\right) = 3 {\left(0\right)}^{2} - 10 \left(0\right) + 5 = 5$

Hence the normal line passes through the point $\left(0 , 5\right)$.

To find the slope of the normal line, first find the slope of the tangent line at $x = 0$, which is equal to the value of the function's derivative at the same point.

To find the derivative, use the power rule.

$f \left(x\right) = 3 {x}^{2} - 10 x + 5$

$f ' \left(x\right) = 6 x - 10$

The slope of the tangent line at $x = 0$ is

$f ' \left(0\right) = 6 \left(0\right) - 10 = - 10$

Now, since the normal line and tangent line are perpendicular, their slopes will be opposite reciprocals. The opposite reciprocal of $- 10$ is $\text{1/10}$.

So, we know the normal line passes through $\left(0 , 5\right)$ and has a slope of $\text{1/10}$. Thus, the equation of the normal line is

$y = \frac{1}{10} x + 5$

Graphed are the function and its normal line:

graph{(3x^2-10x+5-y)(y-1/10x-5)=0 [-16.81, 23.74, -5.3, 14.97]}