What is the equation of the normal line of #f(x)=-x^4+4x^3-x^2+5x-6# at #x=2#?

Question

William Abdalla · Accepted Answer

Slope of normal is #x+17y-818=0# As the function is #f(x)=-x^4+4x^3-x^2+5x-6#, the slope of tangent at any point will be the value of #f'(x)# at that point. As #f'(x)=-4x^3+12x^2-2x+5#, the slope of the tangent at #x=2# will be #-4*2^3+12*2^2-2*2+5=-32+48-4+5=17#. And slope of normal would be #-1-:17=-1/17# Note that value of function at #x=2# is #f(x)=2^4+4*2^3-2^2+5*2-6# = #16+32-4+10-6=48# Hence, slope of normal is #-1/17# and it passes through #(2,48)# its equation of normal is #y-48=-1/17(x-2)# or #17(y-48)=-x+2# or #x+17y-818=0#

Evelyn Agard · Answer

undefined To find the equation of the normal line at x=2 for the function f(x)=-x^4+4x^3-x^2+5x-6, we need to determine the slope of the tangent line at x=2 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 at x=2, we take the derivative of the function f(x) with respect to x and evaluate it at x=2.

The derivative of f(x) is f'(x)=-4x^3+12x^2-2x+5.

Evaluating f'(x) at x=2, we get f'(2)=-4(2)^3+12(2)^2-2(2)+5=-32+48-4+5=17.

The slope of the tangent line at x=2 is 17.

To find the slope of the normal line, we take the negative reciprocal of 17, which is -1/17.

Now, we have the slope of the normal line and the point (2, f(2)) on the function.

To find the y-coordinate of the point (2, f(2)), we substitute x=2 into the function f(x).

f(2)=-(2)^4+4(2)^3-(2)^2+5(2)-6=-16+32-4+10-6=16.

Therefore, the point (2, f(2)) is (2, 16).

Using the point-slope form of a linear equation, y-y1=m(x-x1), where m is the slope and (x1, y1) is a point on the line, we can write the equation of the normal line.

Plugging in the values, we have y-16=(-1/17)(x-2).

Simplifying the equation, we get y=(-1/17)x+(34/17).

Hence, the equation of the normal line of f(x)=-x^4+4x^3-x^2+5x-6 at x=2 is y=(-1/17)x+(34/17).