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). 

2 Answers | Add Yours

Top Answer

embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.

kikinc4's profile pic

kikinc4 | Student, Grade 11 | (Level 1) Honors

Posted on

Given `` 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:

`` so ``

The normal line is perpendicular to the tangent so has a slope of `` .

The equation of the normal line has slope `` and contains point (-2,-2) so `` or `` is the equation of the normal line.

When x=-10 we have `` so the point (-10,0) lies on the normal line.

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

YOU'RE THE BEST

We’ve answered 318,919 questions. We can answer yours, too.

Ask a question