Data la tunzione $f: R \rightarrow \mathbb{R}$ tale che $y=-\frac{1}{2} x^2+2 x+3$
a. calcola $f(-2), f\left(-\frac{1}{2}\right), f(0), f\left(\frac{1}{4}\right)$
b. trova $4f \left(2^{-1}\right)-\frac{1}{4} f(2)$
Data la tunzione $f: R \rightarrow \mathbb{R}$ tale che $y=-\frac{1}{2} x^2+2 x+3$
a. calcola $f(-2), f\left(-\frac{1}{2}\right), f(0), f\left(\frac{1}{4}\right)$
b. trova $4f \left(2^{-1}\right)-\frac{1}{4} f(2)$
Foto dritta!
y = - 1/2·x^2 + 2·x + 3
y = - 1/2·(-2)^2 + 2·(-2) + 3-----> y = -3
y = - 1/2·(- 1/2)^2 + 2·(- 1/2) + 3----> y = 15/8
y = - 1/2·0^2 + 2·0 + 3----> y = 3
y = - 1/2·(1/4)^2 + 2·(1/4) + 3----> y = 111/32
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y = - 1/2·(2^(-1))^2 + 2·2^(-1) + 3----> y = 31/8
y = - 1/2·2^2 + 2·2 + 3----> y = 5
4·(31/8) - 1/4·5 = 57/4