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]}

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