Question

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

Answer 1

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Answer:

See Explanation

Step-by-step 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:

$f"(x) = (-x,ay)$

Lastly: Shifted up

Let the units up be b

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

So, the new function becomes:

$f"'(x) = (-x,ay+b)$

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

$f(x) = x^3$

Vertical compress $f(x) = x^3$ by $\frac{1}{2}$

$f'(x) = \frac{1}{2}x^3$

Reflect $f'(x) = \frac{1}{2}x^3$ over y-axis

Rule: $f"(x)= f'(-x)$

Solving f'(-x)

$f'(x) = \frac{1}{2}x^3$

$f'(-x) = \frac{1}{2}(-x)^3$

$f'(-x) = -\frac{1}{2}x^3$

So:

$f"(x) = -\frac{1}{2}x^3$

Lastly: Shift $f"(x) = -\frac{1}{2}x^3$ by 3

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

$f'"(x) = f"(x + 3)$

Solving: f"(x + 3)

$f"(x) = -\frac{1}{2}x^3$

Substitute x + 3 for x

$f"(x + 3) = -\frac{1}{2}(x + 3)^3$

So:

$f'"(x) = f"(x + 3)$

$f"'(x) = -\frac{1}{2}(x + 3)^3$