How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?

Given: $\displaystyle y=x^2+2x$

Any help is welcomed :)

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- May 30th 2009, 10:07 AMhemiPoints on a graph (tangent line finding)
How can I determine the point(s), if any, at which the graph of the function has a horizontal tangent line?

Given: $\displaystyle y=x^2+2x$

Any help is welcomed :) - May 30th 2009, 10:36 AMursa
hi hemi

differentiate your function

you will get dy/dx. it is the slope of the tangent.

since in the question it has asked for horizontal line,means tangent parellel to x axis.

therefore, its slope will be zero.

put dy/dx=0

and get the value of x

put this value in the function and get the value of y.

thank you - May 30th 2009, 01:15 PMhemi
- May 30th 2009, 01:33 PMursa
hi hemi

this is your solution

y=2*x+x^2

dy/dx=2+2*x. this is the slope of the tangent.

since tangent is parellel to x-axis therefore

dy/dx=0

=> 2+2*x=0

=> x=-1

put this value in function to get the value of y

=> y=-1

therefore, the point on the function at which tangent of the function will be horizontal is (-1,-1).

Thank You - May 31st 2009, 07:16 AMHallsofIvy
You could have done this, by the way, without using Calculus. This is a quadratic function so its graph is a vertical parabola. A tangent line will be horizontal only at its vertex. And we can get the vertex by completing the square: $\displaystyle y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+1)^2- 1$. For x= -1 we have $\displaystyle y= 0^2+ 1= 1$. For any other value of x, x+ 1 is NOT 0 so $\displaystyle (x+ 1)^2$ is positive and y is greater than 1. (-1, 1) is the vertex.