# Thread: transfromations of an exponential function??

1. ## transfromations of an exponential function??

Consider the graph of y = ex. (a) Find the equation of the graph that
results from reflecting about the line y = 8.

y = 1
_________________________

(b) Find the equation of the graph that results from reflecting about
the line x = 2.

y =_____________________________

2
How would i solve this, and can u show the steps?

2. Hello Sneaky
Originally Posted by Sneaky
Consider the graph of y = ex. (a) Find the equation of the graph that
results from reflecting about the line y = 8.
Suppose that the point $(x_1, y_1)$ is transformed into the point $(x_2, y_2)$, by a reflection in the line $y = 8$. Then, since the reflection line is horizontal, the x-coordinate doesn't change, and so:

$x_2 = x_1$ (1)

The fundamental property of a reflection is that the distances of a point and its reflection from the mirror-line are equal. Let's assume (and it won't make any difference if it's the other way around) that $y_1 < 8$ and $y_2 > 8$. Then these distances are $(8-y_1)$ and $(y_2-8)$, respectively. So

$8-y_1 = y_2-8$

$\Rightarrow y_2 = 16-y_1$ (2)

So, since $y_1 = e^{x_1}$, the equation relating $y_2$ and $x_2$, when we use (1) and (2), is:

$y_2 = 16 - e^{x_2}$

and so the equation of the reflected graph is

$y = 16-e^x$

(b) Find the equation of the graph that results from reflecting about
the line x = 2.

y =_____________________________

2
How would i solve this, and can u show the steps?
Do this in the same way. Assume that $(x_1,y_1)\rightarrow (x_2,y_2)$. Then, since the mirror-line is vertical

$y_2=y_1$

and this time it's the horizontal distances from the mirror-line that are equal. So

$2-x_1 = x_2-2$

$\Rightarrow x_1 = 4-x_2$

And, once again, $y_1=e^{x_1}$, so how are $y_2$ and $x_2$ related? This will then give you the equation of the transformed graph.

Can you complete this now?