What is the equation of the line normal

Colt Rhodes

Colt Rhodes

Answered question

2022-04-09

What is the equation of the line normal to f(x)=x33x2 at x=4?

Answer & Explanation

ysnlm8eut

ysnlm8eut

Beginner2022-04-10Added 14 answers

We have: f(x)=x33x2
First, let's find the y-coordinate corresponding to the given value of x:
f(4)=(4)33(4)2
f(4)=6448
f(4)=16
So we must find the equation of the normal line at the point (4,16).
Then, let's differentiate f(x):
f(x)=3x26x
For x=4, the slope of the line is:
f(4)=3(4)26(4)
f(4)=4824
f(4)=24
The gradient of the normal line is the reciprocal of this value:
 Gradient of normal line =(f(4))1
 Gradient of normal line =124
Now, let's express the equation of the normal line in point-slope form:
yy1=m(xx1)
y16=124(x4)
y4=124x16
∴=124x236
Therefore, the equation of the normal line is y=124x236

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