What is the equation of the line normal to #f(x)=x^2+8x - 1# at #x=5#?

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Answer 1 · 2016-02-19T23:33:38

#y=-x/18+1157/18#

Let #y=x^2+8x-1#

#y'=2x+8#

This is equal to the gradient of the tangent line #m#.

At #x=5#:

#m=(2xx5)+8=18#

If #m'# is the gradient of the normal line then:

#m'm=-1#

#:.m'=-1/m=-1/18#

and:

#y=5^2+(8xx5)-1=25+40-1=64#

The equation of the normal line is of the form:

#y=m'x+c#

#:.64=(-1/18xx5)+c#

#:.c=64+5/18=1157/18#

So the equation of the normal line is:

#y=-x/18+1157/18#

graph{(y+x/18-1157/18)(x^2+8x-1-y)=0 [-180, 180, -90, 90]}