Quadratic Function 461774

1. **State the problem:** Find a quadratic function $f(x) = ax^2 + bx + c$ such that: - $f(-1) = 2$ - $f'(0) = 0$ - $\int_0^1 f(x) \, dx = -2$ 2. **Recall formulas and rules:** - The derivative of $f(x)$ is $f'(x) = 2ax + b$. - The definite integral of $f(x)$ from 0 to 1 is $\int_0^1 (ax^2 + bx + c) \, dx = \left[ \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx \right]_0^1 = \frac{a}{3} + \frac{b}{2} + c$. 3. **Use the conditions to form equations:** - From $f(-1) = 2$: $$a(-1)^2 + b(-1) + c = a - b + c = 2$$ - From $f'(0) = 0$: $$2a(0) + b = b = 0$$ So, $b = 0$. - From $\int_0^1 f(x) dx = -2$: $$\frac{a}{3} + \frac{b}{2} + c = \frac{a}{3} + 0 + c = -2$$ 4. **Substitute $b=0$ into the first equation:** $$a - 0 + c = 2 \implies a + c = 2$$ 5. **Now solve the system:** $$\begin{cases} a + c = 2 \\ \frac{a}{3} + c = -2 \end{cases}$$ 6. **Subtract the second equation from the first:** $$\left(a + c\right) - \left(\frac{a}{3} + c\right) = 2 - (-2)$$ $$a - \frac{a}{3} = 4$$ $$\frac{3a - a}{3} = 4$$ $$\frac{2a}{3} = 4$$ $$2a = 12$$ $$a = 6$$ 7. **Find $c$ using $a + c = 2$:** $$6 + c = 2$$ $$c = 2 - 6 = -4$$ 8. **Write the final quadratic function:** $$f(x) = 6x^2 + 0 \cdot x - 4 = 6x^2 - 4$$ **Answer:** $$\boxed{f(x) = 6x^2 - 4}$$