Answer:
(See explanation and attachment below)
Step-by-step explanation:
The image is equal to:
[tex]I(x) = \left(\frac{1}{3}\cdot x \right)^{2}[/tex]
[tex]I(x) = \frac{1}{9}\cdot x^{2}[/tex]
Which means that [tex]I(x) = \frac{1}{9}\cdot P(x)[/tex], that is, a scale factor is applied to preimage in order to diminish growth rate of parabola at same values of x.
Some values of P(x) and I(x) are presented below:
x P(x) I(x)
0 0 0
1 1 1/9
2 4 4/9
3 9 1
The following image corresponds to both functions.
