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Cubic equation that has been transformed from the parent function with a vertical compression, reflected over the y-axis and shifted up.

User Rpax
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1 Answer

6 votes

Question is incomplete

Answer:

See Explanation

Explanation:

Let the cubic function be


f(x) = (x,y)

First transformation: Vertical compression

Let the scale factor be a where
0 < |a| < 1

The new function becomes:


f'(x) = (x,ay)

Next: Reflection over y-axis

The rule is:
(x,y)=>(-x,y)

The new function becomes:


f

Lastly: Shifted up

Let the units up be b

The rule is:
(x,y)=>(x,y+b)

So, the new function becomes:


f

Using the rules states above, assume the cubic function is:


f(x) = x^3

Vertical compress
f(x) = x^3 by
(1)/(2)


f'(x) = (1)/(2)x^3

Reflect
f'(x) = (1)/(2)x^3 over y-axis

Rule:
f

Solving f'(-x)


f'(x) = (1)/(2)x^3


f'(-x) = (1)/(2)(-x)^3


f'(-x) = -(1)/(2)x^3

So:


f

Lastly: Shift
f by 3

Rule:
(x,y)=>(x,y+b)


f'

Solving: f"(x + 3)


f

Substitute x + 3 for x


f

So:


f'


f

User Cryptc
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