Posted by Melinda on Saturday, December 10, 2011 at 12:16pm.
Can someone help me with this question? I got the transformation for horizontal shift and vertical shift, however, I don't know how to find the vertical/horizontal compression/stretch.
1. Describe the transformations that were applied to the parent function to create the graph shown below. Then write the equation of the transformed function.
Parent function y=x^4
Here is the link to graph:
imgur dot com/Gkaln

Precalculus  Reiny, Saturday, December 10, 2011 at 12:37pm
I looked your graph and the new vertex is (1,4)
so y = x^4 must have been translated to
y = (x+1)^4 + 4
but there could also be a compression , so let the new curve be
y = a(x+1)^4 + 4
from the graph we can see that (0,1) lies on the new curve
1 = a(1^4) + 4
3  a
so we have y = 3(x+1)^4 + 4
test for the other point shown
let x = 2
y = 3(1)^4 + 4
= 1
all looks good

Precalculus  Reiny, Saturday, December 10, 2011 at 12:38pm
8th line should have said
3 = a
(but you probably guessed that was a typo)

Precalculus  Melinda, Saturday, December 10, 2011 at 12:39pm
So it is a vertical stretch by 3?

Precalculus  Melinda, Saturday, December 10, 2011 at 12:43pm
How did you find out what a is if you knew it was UP 4, left 1?

Precalculus  Reiny, Saturday, December 10, 2011 at 2:37pm
Yes, the maginitude of the stretch is 3,
the  tells me the curve is opening downwards
I knew since the vertex was (1,4)
and the parent graph of y = x^4 has a vertex at (0,0)
the graph must have moved 1 unit to the left and 4 units up
so we would get
y4 = (x+1)^4
moving the 4 to the right made it +4
but y = (x+1)^4 + 4
would not pass through (0,1) as your picture shows.
so there must have been a stretch/compression.
that would be caused by some number in front of
(x+1)^4
so that is why I chose y = a(x+1)^4 + 4
subbing in the point (0,1) gave me a value of
a = 3
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