Respuesta :

sqdancefan

Answer:

See below.

Step-by-step explanation:

The one that looks like the attachment.

It is vertically shortened from f(x) = x^2, so appears wider than f(x) = x^2 does. The vertex is still at (0, 0), it still opens upward, and it remains symmetrical about the y-axis.

Ver imagen sqdancefan
gmany

[tex]f(x)=\dfrac{1}{2}x^2\\\\for\ x=0\to f(0)=\dfrac{1}{2}(0)=0\to(0,\ 0)\\\\for\ x=\pm1\to f(\pm1)=\dfrac{1}{2}(\pm1)^2=\dfrac{1}{2}(1)=\dfrac{1}{2}\to\left(-1,\ \dfrac{1}{2}\right);\ \left(1;\ \dfrac{1}{2}\right)\\\\for\ x=\pm2\to f(\pm2)=\dfrac{1}{2}(\pm2)^2=\dfrac{1}{2}(4)=2\to(-2,\ 2);\ (2,\ 2)\\\\for\ x=\pm4\to f(\pm4)=\dfrac{1}{2}(\pm4)=\dfraC{1}{2}(16)=8\to(-4,\ 8);\ (4,\ 8)[/tex]

Ver imagen gmany

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