What is the equation of the normal line of f(x)=5x^4+3x^3+x^2-2x-4 at x=1?

1 Answer
Jan 11, 2016

29y+x-88=0

Explanation:

Differentiate the function to find its gradient.
f'(x)=20x^3+9x^2+2x-2
Use the point x=1 to find the gradient of the curve at that point.
f'(1)=20+9+2-2=29
The gradient of the normal is the negative reciprocal of the gradient of the tangent at a point.
So the gradient of the normal at x=1 is m=-1/29
To find the coordinates of the point the normal passes through, substitute x=1 into f(x)
f(1)=5+3+1-2-4=3
So coordinates are (x_1, y_1)=>(1, 3)
Equation of a line is: y-y_1=m(x-x_1)
y-3=-1/29(x-1)

29y-87=-(x-1)

29y+x-88=0