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

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Answer 1 · 2016-05-09T11:16:24

$x - 17 y = 880$ $x-17y=880$

The slope of an equation is given by its derivative.

If $f (x) = x^{3} - 7 x^{2} - x$ $f(x)=x^3-7x^2-x$
then $f'(x)=3x^2-14x-1$
and
at $x=4$
the slope is $m= f'(4)=3(4^2)-14(4)-1=-17$

If a line has a slope of $m$ then any line perpendicular to it (i.e. its "normal") has a slope of $-1/m$

At $x=4$
$color(white)("XXX")f(4)=4^3-7(4^2)-4 = -52$

So the required normal line
$color(white)("XXX")$ has a slope of $1/17$
and
$color(white)("XXX")$ passes through the point $(4,-52)$

Using the slope-point form:
$color(white)("XXX")y-(-52)=1/17(x-4)$

$color(white)("XXX")rarr 17y+884 = x-4$

or, in standard form:
$color(white)("XXX")x-17y=880$
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What is the equation of the line normal to f(x)=x^3 - 7x^2-x f(x)=x3−7x2−xf(x)=x^3 - 7x^2-x at x=4x=4x=4?