Question

What is the equation of the line normal to `f(x)=-(x+6)(x+3)+4x^2-8x+2` at `x=0`?

Answer 1

`y=-1/17x-16`

Explanation:

To find the slope of the normal line, we need to differentiate to get `f'(x)` , then plug in zero to find slope of tangent line at `x=0`, then finally find the opposite reciprocal .

`f(x)=-(x+6)(x+3)+4x^2-8x+2`

Differentiate using the product rule and power rule:
`f'(x)=-[(x+6)(1)+(1)(x+3)]+8x-8`

`f'(x)=-x-6-x-3+8x-8`

`f'(x)=6x-17`

Plug in zero:
`f'(0)=6(0)+17=17`

Find opposite reciprocal of tangent line slope to get normal line slope:
`"Slope of normal line"=-1/17`

Equation of normal line (point-slope form):
`f(0)=-16`
`y+16=-1/17(x-0)`

`y=-1/17x-16`