# vertical distance between two parabola

by icosane
Tags: distance, parabola, vertical
 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.
 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?
 Sci Advisor HW Helper Thanks P: 24,975 Yes, I think it does.
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
 P: 2 oh i forgot... a = 1 b = -k c = 25 x2 = x squared tnx tnx ^__^__^

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