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

##### 1 Answer

#### Explanation:

First simplify the function.

#f(x)=-(x^2+2x-3)+4x^2-8x+2#

#f(x)=3x^2-10x+5#

Before finding the normal line at

#f(0)=3(0)^2-10(0)+5=5#

Hence the normal line passes through the point

To find the slope of the normal line, first find the slope of the tangent line at

To find the derivative, use the **power rule**.

#f(x)=3x^2-10x+5#

#f'(x)=6x-10#

The slope of the tangent line at

#f'(0)=6(0)-10=-10#

Now, since the normal line and tangent line are perpendicular, their slopes will be opposite reciprocals. The opposite reciprocal of

So, we know the normal line passes through

#y=1/10x+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]}