Respuesta :
Answer:
y will increase if x is increased,
y will decrease if x is decreased.
Step-by-step explanation:
1/F=1/X+ 1/Y
1/F-1/X=1/Y
X-F/FX=1/Y
FX/X-F=Y
Rearranging, Y=FX/X-F
So this is the function in terms of X, and F is just a constant
Therefore, the function becomes Y=65X/X-65
Explanation:
y = focusing distance
x = distance from lens
Focusing distance y will increase if the object x moves far away from the lens i.e. x is increased,
Similarly, focusing distance y will decrease if the object x moves closer and closer to the lens i.e. x is decreased.
An asymptote is a function that mimics a curve f(x) as x approaches infinity.
A vertical asymptote, however, cannot be crossed in a function. Remember that a function cannot have multiple y values for a given x value, hence the vertical line test for a function. If a function crossed a vertical asymptote and then went back to it, then it would have to go back over itself as it becomes arbitrarily close to the asymptote. Therefore, it is not possible to cross a vertical asymptote.
The focal point, F, the object distance from the lens, X, and the image distance from the the lens, Y, vary according to the lens formula
1. The value of the focusing distance, Y, is negative and increases in
magnitude, when X < 65, and positive and decreasing in magnitude when
X > 65, the line X = 65 is a vertical asymptote.
2. When X > 65, the value of Y is positive and increasing as the object
moves closer to the lens. At X = 65, there is no image when 0 ≤ X < 65, the
value of Y is negative and increasing towards 0.
3. A vertical asymptote is a discontinuity, where the function is not defined
Reasons:
Known;
[tex]\dfrac{1}{F} = \dfrac{1}{X} + \dfrac{1}{Y}[/tex]
Where;
F = The focal length
X = Object distance from lens
Y = Location of film = Image distance from lens
(a) Given that F = 65, we have;
[tex]\dfrac{1}{65} = \dfrac{1}{X} + \dfrac{1}{Y}[/tex]
[tex]\dfrac{1}{65} - \dfrac{1}{X} = \dfrac{1}{Y}[/tex]
[tex]Y = \dfrac{65 \cdot X}{X - 65}[/tex]
Given that 65·X > X - 65 When X > [tex]-\dfrac{65}{64}[/tex], we have;
As X is increased, Y is decreases, with the rate of decrease reducing as the
value of X gets larger.
(b) As the object moves closer to the lens, when X > 65, the value of Y
increases, at X = 65, the value of Y is infinity, which is a vertical asymptote,
as the object moves closer, the value of Y becomes negative, with the
value increasing (becoming less negative) and when X = 0, Y = 0
3. It is not possible to cross the asymptote because the function has a
discontinuity at the asymptote, which represent the point where the image
is at infinity, such that giving that the distance of the image increases as X
approaches the asymptote, crossing the asymptote results in the image
located further than infinity (which is impossible).
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