wie lautet die scheitelpunktform?

Question

Hey

wie forme ich diese gleichung: 1/4 x² + 1/8x + 1/2 in die scheitelpunktform um

Answer 1

Hi

f(x)=1/4*(x²+1/2x+2)

f(x)=1/4*(x²+1/2x+0,25²-0,25²+2)

f(x)=1/4*(x+1/4)²-0,25²+2)

f(x)=1/4*( (x+1/4)² +31/16 )

f(x)=1/4*(x+1/4)² + 31/64

Hier zur Veranschaulichung

 ~plot~1/4 x^2 + 1/8x + 1/2~plot~

Answer 2

$$\begin{aligned}f(x) & =\frac{1}{4}x^{2}+\frac{1}{8}x+\frac{1}{2} & & |\,\cdot4\\4\cdot f(x) & =x^{2}+\frac{1}{2}x+2 & & |\,-2\\4\cdot f(x)-2 & =x^{2}+\frac{1}{2}x & & |\,+\left(\frac{1}{4}\right)^{2}\\4\cdot f(x)-2+\left(\frac{1}{4}\right)^{2} & =x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}\\4\cdot f(x)-2+\left(\frac{1}{4}\right)^{2} & =\left(x+\frac{1}{4}\right)^{2}\\4\cdot f(x)-\frac{31}{16} & =\left(x+\frac{1}{4}\right)^{2} & & |\,+\frac{31}{16}\\4\cdot f(x) & =\left(x+\frac{1}{4}\right)^{2}+\frac{31}{16} & & |\,:4\\f(x) & =\frac{1}{4}\left(x+\frac{1}{4}\right)^{2}+\frac{31}{64}\end{aligned}$$