vertical distance between two parabola


by icosane
Tags: distance, parabola, vertical
icosane
icosane is offline
#1
Feb15-09, 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 = -(x-2)^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.
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icosane
icosane is offline
#2
Feb15-09, 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?
Dick
Dick is offline
#3
Feb15-09, 09:49 PM
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Thanks
P: 25,160
Yes, I think it does.

teacher Guye
teacher Guye is offline
#4
Oct11-11, 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
teacher Guye
teacher Guye is offline
#5
Oct11-11, 05:17 AM
P: 2
oh i forgot...

a = 1
b = -k
c = 25
x2 = x squared

tnx tnx ^__^__^


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