# The equation of a curve is y=2x+8/x^2   Show that the normal to the curve at the point (-2,-2) intersects the x-axis at the point (-10,0).

Given `f(x)=2x+8/x^2` show that the normal to the curve at (-2,-2) intersects the x-axis at (-10,0):

The normal will be perpendicular to the tangent to the curve at (-2,-2).

The slope of the tangent line is the first derivative evaluated at x=-2:

`f'(x)=2-16/x^3` so `f'(-2)=4`

The normal line is perpendicular...

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Given `f(x)=2x+8/x^2` show that the normal to the curve at (-2,-2) intersects the x-axis at (-10,0):

The normal will be perpendicular to the tangent to the curve at (-2,-2).

The slope of the tangent line is the first derivative evaluated at x=-2:

`f'(x)=2-16/x^3` so `f'(-2)=4`

The normal line is perpendicular to the tangent so has a slope of `m=-1/4` .

The equation of the normal line has slope `m=-1/4` and contains point (-2,-2) so `y+2=-1/4(x+2)` or `y=-1/4x-5/2` is the equation of the normal line.

When x=-10 we have `y=-1/4(-10)-5/2=5/2-5/2=0` so the point (-10,0) lies on the normal line.

The green line is the tangent and the red line is the normal.

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