Vertical distance between two parabola

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The discussion centers on finding the vertical distance function d(x) between two parabolas, y = x^2 + 6 and y = -(x-2)^2 + 6, leading to the conclusion that d(x) = 2x^2 - 4x + 4, with a minimum value of 2. The participant initially struggled but realized that d(x) is simply the difference between the two curves. They then posed a new question about determining the value(s) of k for which the line y = kx intersects the parabola y = x^2 + 25 at only one point. The approach involves setting the equations equal and applying the discriminant method to find k, concluding that k should be 10 or -10 for a single intercept. The conversation highlights the importance of understanding quadratic equations and their intersections.
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Homework Statement



Write a function for d(x), the vertical distance between the two curves, and find the minimum value of d(x).

Homework Equations



The equation for parabola one is y = x^2 + 6, for parabola two, y = -(x-2)^2 + 6


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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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?
 
Yes, I think it does.
 
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
 
oh i forgot...

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

tnx tnx ^__^__^
 

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