# Points on a graph (tangent line finding)

• May 30th 2009, 10:07 AM
hemi
Points 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 AM
ursa
Quote:

Originally Posted by hemi
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 :)

hi hemi
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 PM
hemi
Quote:

Originally Posted by ursa
hi hemi
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

Hi,

Thanks for your post -- I'm still a bit confused though, the majority of this stuff is new to me - can you show me how I'd go about doing what you said so I can see and try to learn by examples? (This is how I learn best :))

Thanks!
• May 30th 2009, 01:33 PM
ursa
Quote:

Originally Posted by hemi
Hi,

Thanks for your post -- I'm still a bit confused though, the majority of this stuff is new to me - can you show me how I'd go about doing what you said so I can see and try to learn by examples? (This is how I learn best :))

Thanks!

hi hemi
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 AM
HallsofIvy
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.