What is the equation of the line normal to $f(x)= -1/x$ at $x=-3$ ?

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Answer 1 · 2018-03-12T14:53:21

$color(blue)(y=-9x-82/3)$

The line that is normal to $f(x)=-1/x$ at $x=-3$ is perpendicular to the tangent line at $x=-3$ .

We first need to find the gradient of the tangent line at $x=-3$ . We now find the derivative of $f(x)=-1/x$ .

$dy/dx(-1/x)=1/x^2$

Plugging in $x=-3$ :

$1/(-3)^2=1/9$

We know that if two lines are perpendicular, then the product of their gradients is $-1$

Let gradient of the normal be $bbm$ , then:

$m*1/9=-1=>m=-9$

The normal passes through the point with x coordinate $-3$ . We now need a corresponding $y$ coordinate. Plugging in $x=-3$ into $f(x)=-1/x$ :

$-1/(-3)=1/3$

Using point slope form of a line:

$(y_2-y_1)=m(x_2-x_1)$

$y-(1/3)=-9(x-(-3))$

$color(blue)(y=-9x-82/3)$

Graph:

enter image source here

What is the equation of the line normal to f(x)= -1/x f(x)= -1/x at x=-3x=-3?