
#1
Feb1509, 07:34 PM

P: 51

1. The problem statement, all variables and given/known data
Write a function for d(x), the vertical distance between the two curves, and find the minimum value of d(x). 2. Relevant equations The equation for parabola one is y = x^2 + 6, for parabola two, y = (x2)^2 + 6 3. The attempt at a solution The answer in the back of the book is d(x) = 2x^2  4x + 4, with a minimum value of 2. This is from my old algebra 2 trig textbook and I have no teacher to ask for help, as I am doing self study. Any and all help would be very much appreciated. 



#2
Feb1509, 09:26 PM

P: 51

I figured it out, its just d(x) = y1  y2, seems I was just over thinking things... but I have another question :)
Find the Value(s) of k for which the graph y = kx intersects the graph of y = x^2 + 25 in only one point. If I set the equations equal, I get 0 = x^2  kx + 25, and I know perfect squares have only one root, so it seems as though the answer is 10 or 10. Does this seem reasonable? 



#3
Feb1509, 09:49 PM

Sci Advisor
HW Helper
Thanks
P: 25,160

Yes, I think it does.




#4
Oct1111, 05:16 AM

P: 2

vertical distance between two parabola
from 0 = x2  kx + 25, use the discriminant method to find the value of k that will result to a single intercept only. that is
b2  4ac = 0 for one root or one intercept b2  4ac > 0 for two roots or two intercepts b2  4ac < 0 if you wish them to have no intercepts 



#5
Oct1111, 05:17 AM

P: 2

oh i forgot...
a = 1 b = k c = 25 x2 = x squared tnx tnx ^__^__^ 


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