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Question

Consider the graph of y = ex.
(a) Find the equation of the graph that results from reflecting about the line y = 5.
I tried to plug in 5 for x but that is not correct so for example, e^-5 was not correct.

I'm lost as to what the question is asking me.

Answer 1

Take a graph of y=ex. Now draw a straight line at y=5 (so whatever the value of x y=5). Imagine that this second line is a mirror, draw the reflection of the first line. Find the slope of the reflection (the third line).

Answer 2

y = 5 represents a horizontal line that is 5 units above the x-axis.

When they say "reflect" about that line, they mean draw another graph that is like a mirror image of the first graph, where the "mirror" is the y=5 line.

For example, if there is some point on the (original) graph that is 2 units BELOW the y=5 line; then its "reflected image" will be a point that is 2 units ABOVE the y=5 line. Same holds for every point--just make its mirror image on the opposite side of the y=5 line.

More examples:

original "y": 5; mirror "y": 5
original "y": 7; mirror "y": 3
original "y": 0; mirror "y": 10
original "y": 20: mirror "y": -10

Notice the pattern: the sum of the "original" y and the "mirror" y is always 10.

(original "y") + (mirror "y") = 10

Or:

(mirror "y") = 10 – (original "y")

Since "original 'y'" is e^x, we have

(mirror "y") = 10 – e^x

So the equation for the reflected graph is:

y = 10 – e^x

Answer 3

when you reflect e^0 over y = 5, the image of (0,1) becomes (0,9), since the 1st point is 4 below the y=5 line and the 2nd is 4 above. suppose e^1 were exactly 3 (2 below the line). image after reflection would be 7 (2 above the line). what's happening? we're adding the difference between 5 and y to 5: 5 + (5-y) = y', which simplifies to y' = 10 - y. so the equation after reflection is y' = 10 - e^x.

graph all 3 equations to verify.

Answer 4

Y=5 for all of x. What ever x is, y=5.

Past that point y=(-ex). OR y=(-ex)-5.