Find Equation of the Normal Line

Find the equation of the normal line to the curve
y = x^3 - 4x^2 + 5 at x = 2.

P. S. Make up a similar problem for me to do on my own.

Thank you.

 

Find the equation of the normal line to the curve
y = x^3 - 4x^2 + 5 at x = 2.

We need the negative reciprocal of slope m and the point x = 2

Solve for the slope m

m=y'(2)

y '= 3x^2 - 8x
y'(2)= 3*2^2 - 8*2
y'(2)= 12 - 16
y'(2)= - 4 => m=-4


y(2) = 2^3 - 4*2^2 + 5
y(2) = -3

Our needed point (x1,y1)=(2,-3)

use slope point formula

y -y[1]=-(1/m)(x-x[1]).....plug in given point and slope
y -(-3)=-(1/-4)(x-2)
y +3=(1/4)x-2/4
y =(1/4)x-1/2-3
y =(1/4)x-7/2-> the equation of the normal line


a similar problem for you:

What is the equation of the normal line of f(x)=x^3-4x^2+7x at x=-2?

 

Thank you. I will solve "a similar problem for you" later on today. Don't forget the other three calculus threads.

 

I first need to find m = f '(-2).

f ' (x) = 3x^2 - 8x + 7

f ' (-2) = 35 = our slope.

I now need f (-2) = y-coordinate of the point I will plug into the point-slope formula.

So, f (-2) = -38.

y - y_1 = m(x - x_1)

y - (-38) = 35(x - (-2))

y + 38 = 35(x + 2)

y + 38 = 35x + 70

y = 35x + 70 - 38

The equation of the normal line to the given curve is

y = 35x + 32.

You say?

 

perfect

 

Thanks. What about my work on finding the equation of the tangent line?