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

Question

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

Cameron Alexander · Accepted Answer

#y=1/7 x+24 6/7#

Given:#" "f(x)=4x^2-3x-2#.......................Equation(1)

#=>f^'(x)=8x-3#

So the gradient of the normal will be #1/f^'(x)#

Thus we have #y=1/f^'(x) x+c #...........................Equation(2) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the point on the curve at "x=3)#

#f(3)=4(3)^2-3(3)-2 =25#

Thus for the point #(x,y)->(3,25)# equation(2) becomes:

#25=1/(8(3)-3) (3)+c#

#c=25-1/7 = 24 6/7#

By substitution Equation(2) becomes:

#y=1/7 x+24 6/7#

Isaiah Allen · Answer

undefined To find the equation of the line normal to the function f(x) = 4x^2 - 3x - 2 at x = 3, we need to determine the slope of the tangent line at x = 3 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we take the derivative of the function f(x) with respect to x.

The derivative of f(x) = 4x^2 - 3x - 2 is f'(x) = 8x - 3.

Evaluating f'(x) at x = 3, we get f'(3) = 8(3) - 3 = 21.

The slope of the tangent line at x = 3 is 21.

To find the slope of the normal line, we take the negative reciprocal of the tangent line's slope.

The negative reciprocal of 21 is -1/21.

Therefore, the equation of the line normal to f(x) = 4x^2 - 3x - 2 at x = 3 is y = (-1/21)(x - 3) + f(3).