Find the vertical translation of y=| x | that "supports" the parabola y=x^2.

[MasonAngus]

Find the vertical translation of y=| x | that "supports" the parabola y=x^2.

Find the vertical translation of y=| x[FONT=&quot] | that "supports" the parabola y=x^2. Hint: You should find c so that y=[/FONT][FONT=&quot]| [/FONT]x[FONT=&quot] |+c just touches y=x^2.
So it basically wants me to find the vertical translation that causes y=[/FONT][FONT=&quot]| [/FONT]x[FONT=&quot] | to be tangent to y=x^2 on either side. How do I go about solving this?[/FONT]

 

[mmm4444bot]

Find the vertical translation of y=| x | that "supports" the parabola y=x^2. Hint: You should find c so that y=| x |+c just touches y=x^2.
So it basically wants me to find the vertical translation that causes y=| x | to be tangent to y=x^2 on either side. How do I go about solving this?

Hello Mason.

This question may be answered with a system of two equations.

Also, because each graph is symmetrical about the y-axis, we can ignore what's happening to the left of the y-axis and work with x+c instead of |x|+c.

One equation in the system is easy to write. If two graphs meet at a point, then the functions are equal at that point.

To write the second equation, realize that the slope of each function is the same at a tangent point.

The first derivative gives us slope, so calculate the derivative of each function, and set them equal to one another. That gives you equation 2.

Solve the system of two equations. :cool:

 

[HallsofIvy]

Find the vertical translation of y=| x[FONT=&quot] | that "supports" the parabola y=x^2. Hint: You should find c so that y=[/FONT][FONT=&quot]| [/FONT]x[FONT=&quot] |+c just touches y=x^2.
So it basically wants me to find the vertical translation that causes y=[/FONT][FONT=&quot]| [/FONT]x[FONT=&quot] | to be tangent to y=x^2 on either side. How do I go about solving this?[/FONT]

As the hint says, you want to find c such that y= |x|+ c is tangent to y= x^2. Because of the symmetry, we can just look at x> 0. That is, we want to find c such that y= x+ c is tangent to y= x^2 at some point (a, a^2). In order to be tangent to y= x^2 at (a, a^2), the line y= x+ c must first pass through (a, a^2)- we must have y= a+ c= a^2. In order to be tangent the slope of the line must be the same as the derivative of the parabola function at that point. The slope of y= x+ c is, of course, 1. The derivative of y= x^2, at x= a, is y'= 2a. We must have 2a= 1.

Solve 2a= 1 and a+ c= a^2 for a and c.