Expert: Paul Klarreich - 10/3/2006

Question
Hi Paul,
My name is MB and I am studying Calculus 1.  It's been awhile since I've worked with calculus, so I'm a little "rusty"

The Problem is I need to find all the points on the curve y=(x+4)(x-5) at which the tangent line is horizontal.

From reading my book, I know how to find the equation for a line tangent, but not the points.  My book doesn't explain it very well.

Thank you!!  

Answer
MB Asks in Category Calculus ...
 
Subject:  Points on a Curve
 
Question:  Hi Paul,
My name is MB and I am studying Calculus 1.  It's been awhile since I've worked with calculus, so I'm a little "rusty"

The Problem is I need to find all the points on the curve y=(x+4)(x-5) at which the tangent line is horizontal.

From reading my book, I know how to find the equation for a line tangent, but not the points.  My book doesn't explain it very well.

Thank you!
................................................
Hi, MB,

I think you have it and just don't realize it, but....

First you want the values of x at which there is a horizontal tangent, which means the slope is zero.  This particular function is a quadratic, so there will be just one.

y = x^2 - x - 20

dy/dx = 2x - 1

Set that = 0 and solve:

2x - 1 = 0
x = 1/2

Now you want the actual point on the graph.  That's x, find y:

y = (1/2)^2 - (1/2) - 20 = 1/4 - 1/2 - 20 = -20 1/4, or  -81/4.

So the (only) point is  (1/2, -81/4)

That's all.