What is the equation of the line normal

Delilah Novak

Delilah Novak

Answered question

2022-04-10

What is the equation of the line normal to f(x)=15x2 at x=-3?

Answer & Explanation

folytonr4qx

folytonr4qx

Beginner2022-04-11Added 14 answers

Explanation:
Slope of the tangent of a curve, at a given point, is given by the value of the first derivative at that point. As normal is perpendicular to tangent, if m is the slope of tangent, slope of normal would be 1m.
To find the point, let us put x=-3 in f(x)=15x2 and
f(3)=15(3)2=14=14.
Hence, we are seeking normal at (3,14)
dfdx=(1(5x2))×(2x)=2x(5x2)2
Hence slope of tangent at x=-3 is
2(3)(5(3)2)2=6(4)2=616=38
and slope of normal is 138=83
and equation of line normal is
(y14)=83(x(3))
or 12y-3=-32x-96 or 32x+12y+93=0
ze2m1ingkdvu

ze2m1ingkdvu

Beginner2022-04-12Added 16 answers

f(x)=15x2 and x0=3.
Find the value of the function at the given point: y0=f(3)=14.
The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function, evaluated at x=x0:M(x0)=1f(x0)
Find the derivative: f(x)=(15x2)=2x(x25)2
Hence, M(x0)=1f(x0)=(x025)22x0
Next, find the slope at the given point.
m=M(3)=83
Finally, the equation of the normal line is yy0=m(xx0)
Plugging the found values, we get that y(14)=8(x(3))3
Or, more simply: y=8x3+314

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