Question

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

Answer 1

`y=1/10x+5`

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 `x=0`, find the point the normal line will intercept:

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

Hence the normal line passes through the point `(0,5)`.

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(x)=3x^2-10x+5`

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

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

`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 `-10` is `"1/10"`.

So, we know the normal line passes through `(0,5)` and has a slope of `"1/10"`. Thus, the equation of the normal line is

`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]}