a) take the inverse of your function and show that the original equation comes back at you.
b) for VA, the denominator is zero, so x-k = 0 is the vertical asymptote
for a HA, I visualize what happens to kx/(x-k) as x -- ∞
I see y = k
c) I solved y = kx/(x-k) with y = x+2k and got
x = ± k√2
put each of those values back into your straight line equation to get the corresponding y's.
take the derivative of the first function using the quotient rule, sub in the points you found to get the slope.
Now that you have the slope and the point, find the straight line equation of the tangent.
(I am assuming you know how to follow those steps)
Consider the function y=(kx)/(x-k)
a. Show that y is symmetric with respect to the line y=x.
b. Write the equations of the horizontal and vertical asymptotes of y.
c. Find the point(s) of intersection of y with the line x-y=-2k, and find the slope of the tangent line at the point(s) of intersection.
Please show step-by-step work in your answers. Thanks in advance!!! :)
2 answers
a)
the reflection matrix about the line y=x is
0 1
1 0
In other words, new x = old y
and new y = old x
So what happens if we do this operation on and x old = a and a y old = b? (a,b)
the original would be:
b = (ka)/(a - k)
find new x and new y if the symmetry exists (b,a)
a = (kb)/(b-k)
If that is true, we have the symmetry.
so solve for b
b a - k a = k b
b (a-k) = k a
b = (ka)/(a-k)
the reflection matrix about the line y=x is
0 1
1 0
In other words, new x = old y
and new y = old x
So what happens if we do this operation on and x old = a and a y old = b? (a,b)
the original would be:
b = (ka)/(a - k)
find new x and new y if the symmetry exists (b,a)
a = (kb)/(b-k)
If that is true, we have the symmetry.
so solve for b
b a - k a = k b
b (a-k) = k a
b = (ka)/(a-k)
1 answer