1. **Problem Statement:** We are given a set of points $(x, f(x))$ and asked to approximate the function $f(x)$ using one of the given forms: a. $f(x) = 15x^2$ b. $f(x) = 15x^3$ c. $f(x) = 15x$ d. $f(x) = 15x^4$ We want to find which function best fits the data points: $x$: -2, -1, 0, 1, 2 $f(x)$: 12.5, 8, 3.8, 21.6, 36.92 2. **Approach:** We will calculate the predicted values for each function at the given $x$ values and compare them to the actual $f(x)$ values to see which is closest. 3. **Calculate predicted values:** - For $f(x) = 15x^2$: $$15(-2)^2 = 15 \times 4 = 60$$ $$15(-1)^2 = 15 \times 1 = 15$$ $$15(0)^2 = 0$$ $$15(1)^2 = 15$$ $$15(2)^2 = 60$$ - For $f(x) = 15x^3$: $$15(-2)^3 = 15 \times (-8) = -120$$ $$15(-1)^3 = 15 \times (-1) = -15$$ $$15(0)^3 = 0$$ $$15(1)^3 = 15$$ $$15(2)^3 = 120$$ - For $f(x) = 15x$: $$15(-2) = -30$$ $$15(-1) = -15$$ $$15(0) = 0$$ $$15(1) = 15$$ $$15(2) = 30$$ - For $f(x) = 15x^4$: $$15(-2)^4 = 15 \times 16 = 240$$ $$15(-1)^4 = 15 \times 1 = 15$$ $$15(0)^4 = 0$$ $$15(1)^4 = 15$$ $$15(2)^4 = 240$$ 4. **Compare predicted values to actual values:** | $x$ | Actual $f(x)$ | $15x^2$ | $15x^3$ | $15x$ | $15x^4$ | |-----|--------------|---------|---------|-------|---------| | -2 | 12.5 | 60 | -120 | -30 | 240 | | -1 | 8 | 15 | -15 | -15 | 15 | | 0 | 3.8 | 0 | 0 | 0 | 0 | | 1 | 21.6 | 15 | 15 | 15 | 15 | | 2 | 36.92 | 60 | 120 | 30 | 240 | 5. **Analysis:** - The $15x^2$ function predicts values closer to the actual $f(x)$ values than the others, especially for $x=1$ and $x=2$. - The $15x^3$ and $15x$ functions produce negative or very different values. - The $15x^4$ function predicts values much larger than actual. 6. **Conclusion:** The best rough approximation of the function $f(x)$ from the given options is: $$f(x) = 15x^2$$