Respuesta :

[tex]\bf \begin{array}{rllll} (6&,&3)&f(x)\\ x&&y\\\\ (-\frac{1}{2}x&,&3)&f(-\frac{1}{2}x)\\\\ (-\frac{1}{2}\cdot 6&,&3)\\\\ (-3&,&3) \end{array}[/tex]

keep in mind that, a negative coefficient to "x", will make the graph reflect over the y-axis.

Answer:

this answer is wrong on plato, trust me

Step-by-step explanation:


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